MEASURE AND INTEGRATION

GESTUR ÓLAFSSON

1. MEASURE AND INTEGRATION

1.1. Measure Spaces. See the book

1.2. Measurable Functions. In this section (X, B) is a fixed measurable space.

Lemma 1. Let f : X → \overline{ℝ} and let α ∈ ℝ. Then the following holds:

1. {x ∈ X | f(x) ≤ α} = ⋂_{n=1}^{∞} {x ∈ X | f(x) < α + 1/n};

2. {x ∈ X | f(x) ≥ α} = ⋂_{n=1}^{∞} {x ∈ X | f(x) > α + 1/n}.

Proof. 1) It is clear that the left hand side is contained in the right hand side. Assume that f(x) < α + 1/n for all n. Then f(x) ≤ lim_{n→∞}(α + 1/n) = α. Hence the sets are equivalent. The proof of (2) is the same.

Lemma 2. Let f : X → \overline{ℝ}. Then the following are equivalent:

1. For all α ∈ ℝ we have {x ∈ X | f(x) < α} ∈ B;

2. For all α ∈ ℝ we have {x ∈ X | f(x) ≤ α} ∈ B;

3. For all α ∈ ℝ we have {x ∈ X | f(x) > α} ∈ B;

4. For all α ∈ ℝ we have {x ∈ X | f(x) ≥ α} ∈ B.

Proof. We proof that ”(1)⇒(2)⇒(3)⇒(4)⇒(1)”. Then the theorem follows.

”(1)⇒(2)”: We have {x ∈ X | f(x) ≤ α} = ∩ {x ∈ X | f(x) < α + 1/n} ∈ B by (1), the above lemma and the fact that a countable intersection of sets in B is again in B.

”(2)⇒(3)”: {x ∈ X | f(x) > α} = X \ {x ∈ X | f(x) ≤ α} ∈ B.

”(3)⇒(4)”: {x ∈ X | f(x) ≥ α} = ⋂_{n=1}^{∞} {x ∈ X | f(x) > α + 1/n} ∈ B.

”(4)⇒(1)”: {x ∈ X | f(x) < α} = X \ {x ∈ X | f(x) ≥ α}.

Definition 1. The function f : X → \overline{ℝ} is called measurable if any one of the statements in the above lemma holds.

We denote by M(X), M^{+}(X), M(X, ℝ), and M^{+}(X, ℝ) the set of measurable functions, nonnegative measurable functions, real valued measurable functions, and nonnegative real valued measurable functions respectively.

Lemma 3. Let f, g : X → ℝ be measurable. Then the following holds: 1. Let c ∈ ℝ. Then the functions cf, f + c are measurable.

2. The functions f · g, f ± g, |f|, f ∧ g and f ∨ g are measurable.

Proof. 1) Assume first that c = 0. Then cf = 0. Hence {x ∈ X | cf(x) < α} = ∅ ∈ B if α ≤ 0 and {x ∈ X | cf(x) < α} = X ∈ B if α > 0. If c > 0. Then

{x ∈ X | cf(x) < α} = {x ∈ X | f(x) < α/c} ∈ B

and if c < 0 then

{x ∈ X | cf(x) < α} = {x ∈ X | f(x) > α/c} ∈ B

We have {x ∈ X | f(x) + c < α} = {x ∈ X | f(x) < α − c} ∈ B. Hence f + c is measurable.

2) Let us first show that f + g is measurable. Let α ∈ ℝ. Then we claim that

{x ∈ X | f(x) + g(x) < α} = ⋃_{r∈ℚ} {x | f(x) < r} ∩ {x | g(x) < α − r} ∈ B

To see this assume first that f(y) + g(y) < α. Then there is a r ∈ ℚ such that

f(y) < r < α − g(y)

Hence y ∈ {x | f(x) < r} ∩ {x | g(x) < α − r}. If f(x) < r and g(x) < α − r then obviously f(x) + g(x) < r + (α − r) = α. Hence the claim. It follows that f + g is measurable.

If α ≤ 0 then {x | f(x)^{2} < α} = ∅ ∈ B. For α > 0 we have

\{x | f(x)^{2} > α\} = \{x | f(x) < −\sqrt{α}\} ∪ \{x | f(x) > \sqrt{α}\} ∈ B

Hence f^{2} is measurable. But then

fg = \frac{1}{2} (f + g)^{2} − f^{2} − g^{2}

is also measurable.

Define the function h : X → ℝ by h(x) = 0 if f(x) = 0 otherwise h(x) =sign(f(x)). Then

{x | h(x) < α} = {x | f(x) < 0} ∈ B if a ≤ 0
{x | h(x) < α} = {x | f(x) ≤ 0} ∈ B if 0 < α ≤ 1

and

{x | h(x) < α} = {x | f(x) < 0} ∈ B if 1 < α

Hence h is measurable. It follows that |f| = hf is measurable. Hence the functions

f ∧ g = \frac{1}{2} ((f + g) − |f − g|)

and

f ∧ g = \frac{1}{2} ((f + g) + |f − g|)

are measurable.

Lemma 4. Let {f_{n}}_{n=1}^{∞} be a sequence of real-valued measurable functions. Then the functions

sup f_{n}, inf f_{n}, \overline{lim}f_{n} and \underline{lim}f_{n}

are also measurable.

Proof. Define g(x) = sup f_{n}(x). Let α ∈ ℝ. If g(x) > α. Then there exists at least one n ∈ ℕ such that f_{n}(x) > α. On the other hand if f_{n}(x) > α for one n then g(x) > α. Hence

{x ∈ X | g(x) > α} = ⋃_{n=1}^{∞} {x ∈ X | f_{n}(x) > α} ∈ B .

As inf f_{n} = − sup(−f_{n}) it follows that inf f_{n} is measurable. We have

\overline{lim}f_{n} = inf_{k} [sup_{n≥k} f_{n}]

and

\underline{lim}f_{n} = sup_{k} [ inf_{n≥k} f_{n}]

which are measurable by the first part.

Suppose that (X, B, μ) is a measure space. Let f and g be measurable functions. Then

{x | f(x) ≠ g(x)} = {x | (f − g)(x) ≤ 0} ∩ {x | (f − g) (x) ≥ 0} ∈ B

We say that f = g almost every where (a.e.) if

μ {x | f(x) ≠ g(x)} = 0

Lemma 5. If μ is a complete measure and f is a measurable functions, then f = g a.e. implies that g is measurable.

2. Simple Functions

In this section (X, B, μ) is a measure space. A simple function on X is a function φ : X → ℝ of the form

φ = ∑_{i=1}^{n} c_{i}χ_{E_{i}}

where the sets E_{i} ⊂ X are measurable, χ_{E_{i}} denotes the characteristic function of E_{i} and c_{i} ∈ ℝ. We call the above representation of φ standard if the sets E_{i} are disjoint.

Proposition 1. Let f be a nonnegative measurable function on X. Then there is a sequence {φ_{n}} of simple functions such that

1. φ_{n+1} ≥ φ_{n};

2. f(x) = lim_{n→∞} φ_{n}(x) for each x ∈ X.

If X is σ-finite, then we can choose the functions φ_{n} such that each of them vanishes outside a set of finite measure.

Proof. For n ∈ ℕ and k ≤ n2^{n} let F_{n} := {x | f(x) ≥ n} ∈ B and let

E_{n,k} : = {x | (k − 1)2^{−n} ≤ f(x) < k2^{−n}}
= {x | (k − 1)2^{−n} ≤ f(x)} ∩ {x | f(x) < k2^{−n}} ∈ B

Define

φ_{n} = ∑_{k=1}^{n2^{n}} \frac{k − 1}{2^{n}}χ_{E_{n,k}} + nχ_{F_{n}}

Then φ_{n} ≤ f for all n.

Next we show that φ_{n+1} ≥ φ_{n}: Assume that x ∈ F_{n}. Then φ_{n}(x) = n. If f(x) ≥ n + 1 then φ_{n+1}(x) = n + 1 > φ_{n}(x): If n ≤ f(x) < n + 1, then there exists a k ∈ [n2^{n+1} + 1, (n + 1)2^{n+1}] ∩ ℕ such that

(k − 1)2^{−(n+1)} ≤ f(x) < k2^{−(n+1)}

Hence x ∈ E_{n+1,k} and

φ_{n+1}(x) = \frac{k − 1}{2^{n+1}} ≥ \frac{n2^{n+1}}{2^{n+1}} = n = φ_{n}(x) .

Assume now that x ∈ E_{n,k}. Then

(k − 1)2^{−n} ≤ f(x) < k2^{−n} .

But

\frac{k − 1}{2^{n}} = \frac{(2k − 1) − 1}{2^{n+1}}

and

\frac{k}{2^{n}} = \frac{2k}{2^{k+1}}

Hence for k′ = 2k − 1 or k′ = 2k we have

\frac{k′ − 1}{2^{n+1}} ≤ f(x) < \frac{k′}{2^{n+1}}

It follows that

φ_{n+1}(x) = (k′ − 1)2^{−(n+1)} ≥ (2k − 2)2^{−(n+1)} = k2^{−n} = φ_{n}(x) .

φ_{n}(x) ↗ f(x): Let us first assume that f(x) = ∞. Then x ∈ F_{n} for all n. Hence φ_{n}(x) = n ↗ ∞. Assume now that f(x) < ∞. Then φ_{n}(x) is a monotone increasing sequence and hence convergent. We have to show that lim_{n} φ_{n}(x) = f(x). Let ϵ > 0 be given. Choose N ∈ ℕ such that 2^{−N} < ϵ. Let n ≥ N. Let k be such that x ∈ E_{n,k}. Then

0 ≤ f(x) − φ_{n}(x) < k2^{−n} − (k − 1)2^{−n} = 2^{−n} ≤ 2^{−N} < ϵ

Theorem 1. Let f : X → \overline{ℝ} be measurable. Then there exists simple functions {φ_{n}}_{n} such that

lim_{n→∞} φ_{n}(x) = f(x)

for all x ∈ X.

Proof. Let f_{+} := f ∨ 0 and f_{−} := (−f)_{+}. Then f = f_{+} − f_{−}. Choose simple functions {φ_{+,n}}_{n} and {φ_{−,n}} such that

lim_{n} φ_{+,n}(x) = f_{+}(x)

and

lim_{n} φ_{−,n}(x) = f_{−}(x)

Notice that if f(x) = 0 then φ_{+,n}(x) = φ_{−,n}(x) = 0. Let φ_{n}(x) = φ_{+,n}(x) − φ_{−,n}(x). Then φ_{n} is a simple function and

lim_{n} φ_{n}(x) = f_{+}(x) − f_{−}(x) = f(x) .

3. Integration

Let φ = ∑ c_{i}χ_{E_{i}} be a simple measurable function. Assume that the representation is standard. If φ ≥ 0 define

∫ φ(x) dμ(x) := ∑ c_{i}μ(E_{i})

We sometimes simplify the notation by writing ∫ φ(x) dμ(x) = ∫ φ dμ = ∫ φ. We need to assume that φ ≥ 0 to be sure that the sum on the right hand side exists, because otherwise we might end up with meaningless symbols as ∞ − ∞. We also need to make sure that the integral is well defined, i.e., that the sum on the right hand side does not depend on the representation of φ.

Lemma 6. Let φ = ∑_{i=1}^{n} c_{i}χ_{E_{i}} = ∑_{j=1}^{m} d_{j}χ_{F_{j}}. Then

∑_{i=1}^{n} c_{i}μ(E_{i}) = ∑_{j=1}^{m} d_{j}μ(F_{j})

Proof. We notice first that c_{i} = d_{j} if E_{i} ∩ F_{j} ≠ ∅. We can assume that X = ∪E_{i} = ∪F_{j}. Then E_{i} = ˙∪_{j}E_{i} ∩ F_{j} and F_{j} = ˙∪_{i}E_{i} ∩ F_{j}. Hence

∑_{i=1}^{n} c_{i}μ(E_{i}) = ∑_{i=1}^{n} c_{i} μ(∪_{j}(E_{i} ∩ F_{j}))

= ∑_{i=1}^{n} ∑_{j=1}^{m} c_{i} μ(E_{i} ∩ F_{j})

= ∑_{j=1}^{m} ∑_{i=1}^{n} d_{j} μ(E_{i} ∩ F_{j})

= ∑_{j=1}^{m} d_{j} μ(F_{j})

Lemma 7. Let φ and ψ be nonnegative simple functions. Let c∈ ℝ^{+}. Then

∫ (cφ + ψ) dμ = c ∫ φ dμ + ∫ ψ dμ

Proof. Let φ = ∑_{i} c_{i}χ_{E_{i}} and ψ = ∑_{j} d_{j}χ_{F_{j}} be a standard representation of φ respectively ψ. We can assume that X = ∪E_{i} = ∪F_{j}. (Otherwise we add X \ ∪E_{i} with the constant 0 to the standard representation of φ, and similarly for ψ.) Then cφ = ∑_{i,j}(cc_{i}) χ_{E_{i}∩F_{j}} and ∑_{i,j} d_{j} χ_{F_{j}∩E_{i}} are also standard representations. Furthermore

cφ + ψ = ∑_{i,j}(cc_{i} + d_{j})χ_{E_{i}∩F_{j}}

Hence

∫ cφ + ψ dμ = ∑_{i,j}(cc_{i} + d_{j}) μ(E_{i} ∩ F_{j})
= c ∑_{i} c_{i} ∑_{j} μ(E_{i} ∩ F_{j}) + ∑_{j} d_{j} ∑_{i} μ(E_{i} ∩ F_{j})
= ∑_{i} c_{i}μ(E_{i}) + ∑_{j} d_{j}μ(F_{j})

= c ∫ φ du + ∫ ψ dμ

Definition 2. Let f ∈ M^{+}(X). We define the integral ∫ f(x) dμ(x) of f with respect to μ by

∫ f(x) dμ(x) := sup_{φ} ∫ φ(x) dμ(x)

where the supremum is taken over all simple functions φ ≤ f. If E ∈ B then

∫_{E} f(x) dμ(x) := ∫ f(x)χ_{E}(x) dμ(x)

We will also use the following notations for the integral

∫_{E} fdμ = ∫_{E} f = ∫_{E} f(x) dμ(x)

If E = X then we also write ∫ fdμ instead of ∫_{X} f du.

Lemma 8. Let f, g ∈ M^{+}(X) and E, F ∈ B.

1. If f ≤ g then ∫ f dμ ≤ ∫ g dμ;

2. If E ⊂ F then ∫_{E} f dμ ≤ ∫ g dμ.

Proof. 1) If φ is a simple function with φ ≤ f. Then φ ≤ g.

2) We have fχ_{E} ≤ fχ_{F}.

Corollary 1. Let f ∈ M^{+}(X). Then ∫ fdμ = 0 if and only if f(x) = 0 a.e.

Proof. Assume first that f = 0 a.e. Let φ = ∑ c_{j}χ_{E_{j}} be a simple functions φ ≤ f. Then if c_{j} ≠ 0 it follows that μ(E_{j}) = 0. Hence ∫ φ = 0. It follows that ∫ fdμ = 0.

Assume now that ∫ fdμ = 0. Let

X_{n} := {x | f(x) > 1/n} .

Then X_{n} ⊂ X_{n+1}, μ(X_{n}) ≤ n ∫_{X_{n}} fdμ ≤ n ∫_{X} dμ = 0. Furthermore

⋃_{n=1}^{∞} X_{n} =: X_{∞} = {x | f(x) > 0}

Hence μ(X_{∞}) = μ(∪X_{n}) ≤ ∑ μ(X_{n}) = 0.

Theorem 2 (Monotone Convergence). Let {f_{n}} be a monotone increasing sequence of functions in M^{+}(X) which converges to f, then

lim_{n→∞} ∫ f_{n}(x) dμ(x) = ∫ f(x) dμ(x)

Proof. As f = lim f_{n} it follows that f ∈ M^{+}(X). We have f_{n} ≤ f_{n+1} ≤ f. Hence

0 ≤ ∫ f_{n}dμ ≤ ∫ f_{n+1}dμ ≤ ∫ fdμ.

Thus

lim_{n→∞} ∫ f_{n} dμ ≤ ∫ f dμ

For the other direction it is enough to show that if φ is a simple function such that 0 ≤ φ ≤ f, then ∫ φ ≤ lim ∫ f_{n}. This is because the integral of f is just the supremum over the integral of all such φ. Let 0 < α < 1. Let φ be a simple function with 0 ≤ φ ≤ f. Let

X_{n} := {x ∈ X | f_{n}(x) ≥ αφ(x)}

Then X_{n} is measurable, X_{n} ⊂ X_{n+1} and ∪X_{n} = X. Hence

∫_{X_{n}} αφ dμ ≤ ∫_{X_{n}} f dμ ≤ ∫_{X} f dμ

and - because φ is a simple function -

lim_{n→∞} ∫_{X_{n}} αφ = α ∫_{X} φ dμ ≤ lim_{n→∞} ∫ f_{n} dμ

As α was arbitrary it follows that

sup ∫_{X} φ dμ ≤ lim_{n→∞} ∫ f_{n} dμ

which is what we had to show.

Corollary 2. Let f, g ∈ M^{+}(X) and let c ∈ ℝ^{+}. Then

∫ cf + g dμ = c ∫ f dμ + ∫ g dμ

Proof. Let φ_{n} be a monotone sequence of simple functions converging to f and let ψ_{n} be a monotone increasing sequence of simple functions converging to g. Then cφ_{n}+ψ_{n} is a monotone increasing sequence of simple functions converging to cf +g. Hence:

∫ cf + g dμ = lim [∫ cφ_{n} + ψ_{n} dμ]
= lim [c ∫ φ_{n}dμ + ∫ ψ_{n} dμ]
= c lim ∫ φ_{n}dμ + lim ∫ ψ_{n}dμ
= c ∫ fdμ + ∫ gdμ

Theorem 3 (Fatou’s Lemma). Let {f_{n}} be a sequence in M^{+}(X). Then

∫_{X} \underline{lim}(f_{n}) dμ ≤ \underline{lim} ∫_{X} f_{n} dμ

Proof. Let

g_{m} := inf_{k≥m} {f_{k}} .

Then \{g_{m}\}_{m} is an increasing sequence converging to \underline{lim}(f_{n}). Hence by the Monotone Convergence Theorem

lim ∫ g_{m} dμ = ∫_{X} \underline{lim}(f_{n}) dμ

But if n ≥ m then g_{m} ≤ f_{n}. Hence

∫_{X} g_{m} dμ ≤ ∫_{X} f_{n} dμ, n ≥ m .

It follows that

\underline{lim} ∫_{X} f_{n} dμ = lim_{m→∞} ( inf_{n≥m} ∫_{X} f_{n} dμ)
≥ lim_{m→∞} ∫_{X} g_{m} dμ
= ∫_{X} \underline{lim}(f_{n}) dμ

Definition 3. Let μ and ν be two measures defined on B. Then ν is absolutely continuous with respect to μ if μ(E) = 0 implies that ν(E) = 0.

Corollary 3. Let f ∈ M^{+}(X). Then

μ_{f} : E ↦ μ_{f}(E) := ∫_{E} fdμ

is a measure, which is absolutely continuous with respect to μ.

Proof. We only have to show that μ_{f} is σ-additive. Let {E_{n}}_{n} be a sequence of disjoint sets in B. Let F_{m} = ∪_{n=1}^{m}E_{m}. Then

χ_{F_{m}} = ∑_{n=1}^{m} χ_{E_{n}}

and

fχ_{F_{m}} ≤ fχ_{F_{∞}}

Furthermore we have lim fχ_{F_{m}} = fχ_{F_{∞}}. Hence by the Monotone Convergence Theorem and the finite additivity of the integral:

μ_{f}(∪E_{n}) = ∫_{X} fχ_{F_{∞}}
= lim_{m→∞} ∫ fχ_{F_{m}}
= lim_{m→∞} ∑_{n=1}^{m} ∫_{X} fχ_{E_{n}}

= lim_{m→∞} ∑_{n=1}^{m} μ_{f}(E_{n})

= ∑_{n=1}^{∞} μ_{f}(E_{n}) .

Assume that μ(E) = 0. Then fχ_{E} = 0 a.e. Hence ∫ fχ_{E} = μ_{f}(E) = 0.

Lemma 9. Suppose that {f_{n}} is a monotone sequence in M^{+}(X) that converges almost everywhere to a function f ∈ M^{+}(X). Then

lim ∫ f_{n} dμ = ∫ f dμ

Proof. Let M := x | lim f_{n}(x) = f(x)}. Then {f_{n}χ_{M}} is a monotone sequence that converges to fχ_{M}. By the Monotone Convergence Theorem we have

∫ fχ_{M}dμ = lim ∫ f_{n}χ_{M} dμ

On the other hand we have that f = fχ_{M} a.e. and f_{n}χ_{M} = f_{n} a.e. Hence

∫ f = ∫ fχ_{M}  and  ∫ f_{n} = ∫ f_{n}χ_{M}

This implies now the Lemma.

Theorem 4. Let {g_{n}} be a sequence in M^{+}(X). Then

∫ (∑_{n=0}^{∞} g_{n}) dμ = ∑_{n=0}^{∞} ∫_{X} g_{n} dμ

Proof. Let f_{m} := ∑_{n=0}^{m} g_{m}. Then f_{m} ∈ M^{+}(X) is an increasing sequence with lim f_{m} = ∑_{n=0}^{∞} g_{n}. It follows by the Monotone Convergence Theorem that

∫ (∑_{n=0}^{∞} g_{n}) dμ = ∫ lim f_{m} dμ

= lim ∫ f_{m} dμ
= lim ∑_{n=1}^{m} ∫ g_{n} dμ

= ∑_{n=0}^{∞} ∫ g_{n} dμ

Definition 4. A nonnegative measurable function f is integrable over E ∈ B if ∫_{E} f < ∞. A function f : X → \overline{ℝ} is integrable over E if f is measurable and both f^{+} and f^{−} are integrable over E. The integral of f is then defined as

∫_{E} f dμ = ∫_{E} f ^{+} dμ − ∫_{E} f^{−} dμ

The set of integrable functions is denoted by L^{1}(X) = L^{‘}(X, μ)..

Notice that a function is integrable over E if and only if f ∈ L^{1}(E, μ_{E}) where μ_{E}.

Lemma 10. The function f is integrable over E if and only if |f| is integrable over E and in that case |∫ f| ≤ ∫ |f|.

Proof. We have |f| = f^{+} + f^{−}. Hence |f| is integrable if and only if f^{+} and f^{−} are integrable and

∫_{E} |f| = ∫_{E} f^{+} + ∫_{E} f^{−} ≥ |∫_{e} f^{+} − ∫_{E} f^{−}| = |∫_{E} f|

The next lemma show that the integral of f is independent of the representation of f as a difference of positive functions.

Lemma 11. Let f = f_{1} − f_{2} with f_{1}, f_{2} nonnegative. Then f is integrable if f_{1} and f_{2} are integrable and in that case ∫ f = ∫ f_{1} − ∫ f_{2}.

Proof. We have f = f_{1} − f_{2} = f^{+} − f^{−}. It follows that f(x) ≤ f_{1}(x). So in particular it follows that f^{+}(x) ≤ f_{1}(x). Hence f^{+} is integrable if f_{1} is integrable. Similarly it follows that f^{−} is integrable if f_{2} is integrable. Thus f is integrable if f_{1} and f_{2} are integrable. We have f_{1} + f^{−} = f_{2} + f^{+}. Thus

∫ f_{1} + f^{−} = ∫ f_{2} + f^{+}

But we have already seen that the integral is additive for positive functions. Thus

∫ f_{1} + ∫ f^{−} = ∫ f_{1} + ∫ f^{+} .

Hence - as all the integrals are finite- we have

∫ f_{1} − ∫ f_{2} = ∫ f^{+} − ∫ f^{−} = ∫ f

Proposition 2. Let E ∈ B, f, g ∈ M(X) integrable over E, and c ∈ ℝ, Then the following holds.

1. cf + g is integrable over E and ∫_{E} cf + g = c ∫ f + ∫ g (i.e., the space of integrable functions over E is a vector space and ∫ is a linear functional).

2. If h ∈ M(X) and |h| ≤ |f|, then h is integrable.

3. If g ≤ f a.e. then ∫ g ≤ ∫ f.

4. If g = 0 a.e. then ∫ g = 0.

Proof. 1) The claim is obvious if c = 0. Assume that c > 0. Then

cf + g = c(f^{+} + g^{+}) − (f^{−} + g^{−}).

The claim follows by the last lemma.

2) This follows by Lemma 8 and Lemma 10.

3) f − g ≥ 0 (a.e.). Then (f − g)^{−} = 0 almost everywhere. By Corollary 1 it follows that ∫ (f − g)^{−} = 0. Hence

∫ f − g = ∫ (f − g)^{+} ≥ 0

Hence by (1) ∫ f − ∫ g ≥ 0 or ∫ f ≥ ∫ g.

4) Follows by (3).

Remark 1. If f is integrable, then the set where f is infinite is of measure zero.

Theorem 5 (Lebesgue Dominated Convergence Theorem). Let {f}_{n} be a sequence of integrable functions which converges almost everywhere to a real-valued measurable function f. If there exists an integrable function g such that |f_{n}| ≤ g for all n. Then f is integrable and

∫ f dμ = lim_{n→∞} ∫ f_{n} dμ

Proof. We remark first that ∫ |f| is finite because of Fatou’s Lemma. Hence |∫ f| < ∞. Be redefining the functions f_{n} and f on a set of measure zero we can assume that the convergence is everywhere on X. The function f is measurable and f is integrable by proposition 2. As g + f_{n} ≥ 0 it follows by Fatou’s Lemma that

∫ f + ∫ g = ∫ (f + g)
≤ \underline{lim} ∫ (g + f_{n})
= ∫ g + \underline{lim} ∫ f_{n}

Hence ∫ f ≤ \underline{lim} ∫ f_{n}. But we also have g − f_{n} ≥ 0. Hence

∫ g − ∫ f = ∫ (g − f)
≤ \underline{lim} ∫ (g − f_{n})
= ∫ g − \overline{lim} ∫ f_{n}

Hence \overline{lim} ∫ f_{n} ≤ ∫ f ≤ \underline{lim} ∫ f_{n}. It follows that lim ∫ f_{n} exists and is equal to ∫ f.

4. L^{p}-Spaces

For 1 ≤ p < ∞ let L^{p}(X) = L^{p}(X, μ) be the space of measurable functions on X such that |f|^{p} is in L^{1}(X). Thus a measurable function is in L^{p} if and only if ∫ |f|^{p} < ∞. Let

||f||_{p} = \sqrt[p]{∫ |f|^{p} dμ}

We have ||f||_{p} = 0 if and only if f = 0 a.e. We will now show that (L^{p}, ||·||_{p}) satisfies the axioms for a Banach space except that ||·||_{p} is only a pseudonorm.

Theorem 6 (Hölder’s Inequality). If f ∈ L^{p} and g ∈ L^{q} with 1 < p < ∞ and 1/p + 1/q = 1 then fg ∈ L^{1} and

||fg||_{1} ≤ ||f||_{p} ||g||_{q}

Proof. The statement is clear if ||f||_{p} = 0 or ||g||_{q} = 0 because then fg = 0 a.e. Let us therefore assume that the right hand side is > 0. Let α ∈ (0, 1). Define ϕ(t) := αt − t^{α}. Then ϕ′(t) = α − αt^{α−1} = α(1 − t^{α−1}). Hence ϕ′(t) < 0 for 0 < t < 1, ϕ′(t) > 0 for 1 < t and ϕ′(1) = 0. It follows that ϕ(t) > ϕ(1) = α − 1. This implies that

t^{α} ≤ αt + (1 − α), t ≥ 0

Take t = a/b with a and b positive. Then [a/b]^{α} ≤ α[a/b] + (1 − α). Multiply this by b to get

a^{α}b^{1−α} ≤ αa + (1 − α)b

Let now p and q be as in the lemma and let α = 1/p. Let a = |f(x)|^{p} / ||f||_{p}^{p} and let B = |g(x)|^{q} / ||g||_{q}^{q}. Then

\frac{|f(x)g(x)|}{||f||_{p} ||g||_{q}} ≤ \frac{|f(x)|^{p}}{p ||f||_{p}^{p}} + \frac{|g(x)|^{q}}{q ||g||_{q}^{q}}

Integrating over X gives

||fg||_{1} / ||f||_{p} ||g||_{q} ≤ \frac{1}{p} + \frac{1}{q} = 1

or ||fg||_{1} ≤ ||f||_{p} ||g||_{q}.

Theorem 7 (Cauchy-Bunyakovskii-Schwarz Inequality). Let f, g ∈ L^{2} Then fg is integrable and

|∫ fg dμ| ≤ ∫ |fg| dμ ≤ ||f||_{2} ||g||_{2}

Theorem 8 (Minkowski’s Inequality). Let 1 ≤ p < ∞. Then L^{p} is a vector space, i.e., If f, g ∈ L^{p}, and λ ∈ ℝ, then λf + g ∈ L^{p}. Furthermore

||f + g||_{p} ≤ ||f||_{p} + ||g||_{p}

Proof. This is clear for p = 1, so let us assume that 1 < p. Let q be given by 1/p + 1/q = 1. This is the case if and only if p + q = pq, or p = pq − q. That λf ∈ L^{p} is clear. Let x ∈ X. Then

|f(x) + g(x)| ≤ |f(x)| + |g(x)| ≤ 2 max{|f(x)| , |g(x)|}

Hence

|f(x) + g(x)|^{p} ≤ 2^{p} max {|f(x)|^{p} , |g(x)|^{p}} ≤ 2^{p} (|f(x)|^{p} + |g(x)|^{p})

It follows that |f(x) + g(x)|^{p} is integrable. Hence f + g ∈ L^{p}. Now (p − 1)q = p. Hence |f + g|^{p−1} ∈ L^{q} and: We get now from Hölder’s inequality:

|f + g|^{p} = |f + g| |f + g|^{p−1} ≤ |f| |f + g|^{p−1} + |g| |f + g|^{p−1}

Lemma 12. Let {g_{n}}_{n∈ℕ} be a sequence in L^{p}, 1 ≤ p < ∞, such that for all ϵ > 0 there is a N ∈ ℕ such that for all n, m ≥ N we have ||g_{n} − g_{m}|| < ϵ. Then there is a g ∈ L^{p} such that lim_{n} ||g_{n} − g|| = 0.

Proof. To be added.

Let

N = {f ∈ M(X) | f = 0 a.e.}

Let L^{p}(X, μ) := L^{p}(X, μ)/N. Then ||·||_{p} is a well definite nonnegative function on L^{p}. Notice that the elements in L^{p} are classes of functions but all the elements in a class are equal except on a set of measure zero. Because of this will will most of the time just work with the elements of L^{p} as if they were functions. If f ∈ L^{p} then we will sometimes write [f] for the corresponding class. Definition of ||·||_{∞}, L^{∞}(X, μ) and L^{∞}(X, μ) will be added. It follows from what we have done so far that L^{p} is Banach space:

Theorem 9. Let 1 ≤ p < ∞. Then L^{p}(X, μ) is a Banach space.

4.1. Signed Measures. To motivate the following constructions and theorems let us assume that f ∈ L^{1}(X). Define a function

μ_{f} : B → ℝ

by

μ_{f}(E) := ∫_{E} f(x) dμ(x)

Then map μ_{f} has the following properties:

1. μ_{f}(∅) = 0.

2. μ_{f} is σ-additive: Let {E_{n}} be a collection of disjoint measurable sets. Then

μ_{f} (⋃ E_{n}) = ∑ μ_{f}(E_{n})

and the series on the right converges absolutely. This follows from the fact, that with f ∈ L^{1} we also have |f| ∈ L^{1} and

μ_{|f|} (∪E_{n}) = ∑ ∫_{E_{n}} |f| dμ ≥ ∑ |∫_{E_{n}} f dμ| = ∑ |μ_{f}(E_{n})| .

We notice that this is necessary because the left hand side does not depend on the order we number the sets E_{j} but only on the union. Hence the left hand side has to be independent of the ordering that we sum. But that is the same as saying that the sum converges absolutely.

3. The map μ_{f} is not a measure, because it can take negative values. But we have

μ_{f} = μ_{f^{+}} − μ_{f^{−}}

and bot μ_{f^{+}} and μ_{f^{−}} are measures. Thus we can decompose the ”signed measure” μ_{f} into the difference of two measures.

4. Let P := {x | f(x) > 0}, N := {x | f(x) < 0} and Z := {x | f(x) = 0}. Then

X = P ˙∪N ˙∪Z

and

(a) If E ⊂ P then μ_{f}(E) ≥ 0. Thus P is a ”positive set” for μ_{f};

(b) If E ⊂ N then μ_{f}(E) ≤ 0. Thus N is a ”negative set” for μ_{f};

(c) If E ⊂ Z then μ_{f}(E) = 0. Thus Z is a ”null set” for μ_{f}. We would like to generalize this to other set-functions. We start with the fol-

lowing definition:

Definition 5. By a signed measure on the measurable space (X, B) we mean an function μ : B → \overline{ℝ} such that

1. μ assume at most one of the values ∞ and −∞.

2. μ(∅) = 0.

3. Let {E_{j}} be a disjunct collect of sets in B. Then

μ (∪E_{j}) = ∑_{j} μ(E_{j})

where the series on the right hand side converges absolutely.

Definition 6. Let μ be a signed measure. Let A ∈ B. Then we call A for: 1. positive if for all B ⊆ A we have μ(B) ≥ 0.

2. negative if it is positive for −μ.

3. null set if it is both positive and negative.

Notice that A is a null set if and only if μ(B) = 0 for all B ⊆ A. Notice that it is necessary to request each subset to be positive, negative respectively zero. For example take X = [−1, 1] and f(x) =sign(x). Let μ = λ_{f} where λ is the Lebesgue measure on [−1, 1]. Then μ(X) = 0, but X is not a null set. If A = [−0.5, 1] then μ(A) = 0.5 > 0, but μ ([−0.5, 0]) = −0.5 < 0 whereas μ([0, 1]) = 1 > 0. Thus A is not positive.

Lemma 13. Let μ be a signed measure on (X, B). Let E ∈ B be such that 0 < μ(E) < ∞. Then there is a positive set A ⊆ E such that μ(A) > 0.

Proof. If E is positive, then we are done. So let us assume that E is not positive. Then there exists a measurable set E_{1} ⊆ E such that μ(E_{1}) < 0. Let n_{1} ∈ ℕ be the smallest integer such that there exists a set E_{1} ⊆ E with

μ(E_{1}) < −1/n_{1}.

If E \E_{1} is not positive then we can find a smallest integer n_{2} such that there exists a set E_{2} ⊆ E \ E_{1} such that

μ(E_{2}) < −1/n_{2}.

Inductively we construct disjoint measurable sets {E_{k}} such that if E \ ∪_{j=1}^{k}E_{j} is not positive, then n_{k+1} is the smallest integer such that there exists a measurable set E_{k+1} ⊆ E \ ∪_{j=1}^{k}E_{j} and μ(E_{k+1}) ≤ −1/n_{k+1}. Notice that by construction we have

n_{j} ≤ n_{j+1} .

Furthermore

μ(E) = μ(E \ ∪_{j=1}^{k}E_{j}) + ∑_{j=1}^{k} μ(E_{j})

≤ μ(E \ ∪_{j=1}^{k}E_{j}) − ∑_{j=1}^{k} \frac{1}{n_{j}}

In particular 0 < μ(E) ≤ μ(E \ ∪_{j=1}^{k}E_{j}). Notice that if E \ ∪_{j=1}^{k}E_{j} is positive then this implies our statement. Thus we can assume that the sequence {E_{j}} is infinite. Let F := ∪E_{j} and A := E \ F. Then both A and F have finite measures and μ(A) ≥ μ(E) > 0. Both statements follows by μ(E) = μ(A) + μ(F) < ∞ and the fact that μ can only take one of the values ∞ or −∞. But

μ(F) = − ∑ \frac{1}{n_{j}}

Hence ∑ \frac{1}{n_{j}} < ∞. In particular we get that n_{j} is a monotone sequence converging to ∞.

We will now show that A is positive. Let ϵ > 0. Choose N ∈ ℕ such that

\frac{1}{n_{k} − 1} < ϵ ∀k ≥ N

This is possible because n_{k} ≤ n_{k+1}. As A ⊆ E \ ∪_{j=1}^{k}E_{j} it follows that A can not contain a set B such that

μ(B) ≤ −1/n_{k+1} ≤ \frac{−1}{n_{k} − 1} < −ϵ.

Hence μ(B) ≥ −ϵ. As ϵ was arbitrary it follows that μ(B) ≥ 0.

Theorem 10 (Hahn Decomposition Theorem). Let μ be a signed measure on a measurable space (X, B). Then there is a positive set P and a negative set N such that

X = P ˙∪N .

Proof. We can assume that μ(L) < ∞ for all L ∈ B. Otherwise we replace μ by −μ. We notice that the set of positive sets is not empty as ∅ is positive. Let

λ := sup_{P positive} μ(P) ≤ ∞ .

Let {P_{n}} be a sequence of positive sets such that lim μ(P_{n}) = λ, and let P := ∪P_{n} and let F_{1} := P_{1} and

F_{n+1} := P_{n+1} \ ∪_{j=1}^{n}P_{j}

We claim that P is positive. Let E ⊆ P. Then

E = ∪E ∩ P_{n} = ˙∪E ∩ F_{n}

As P_{n} is positive it follows that μ(E ∩ F_{n}) ≥ 0. Hence

μ(E) = ∑ μ(E ∩ F_{n}) ≥ 0

In particular we get

μ(P) ≤ λ.

But on the other hand we have μ(P) ≥ μ(P_{n}) as P_{n} ⊆ P and P is positive. Hence

λ ≥ μ(P) ≥ sup μ (P_{n}) = lim μ(P_{n}) = λ

It follows that

μ(P) = λ .

In particular

0 ≤ λ < ∞ .

Let N = X \ P. Assume that N is not negative. Then there is E ⊆ N such that μ(E) > 0. The set E can not be positive because then P ∪ E is positive and

λ ≥ μ(P ∪ E) = μ(P) + μ(E) = λ + μ(E) > λ .

But by the last lemma we can find a positive set A ⊆ E such that μ(E) > 0. That again would give a contradiction. Hence N is negative. The theorem follows now because P and N are disjoint by construction.

Notice that the Hahn-decomposition is not unique. For example in the case μ_{f} then P and P ∪Z are positive sets and the negative sets in the Hahn decomposition are N ∪ Z respectively N. But the restriction of μ to P and N is unique:

Lemma 14. Let (X, B, μ) be as in the last theorem. Let X = P_{1} ˙∪N_{1} = P_{2} ˙∪N_{2} be two decompositions of X into positive and negative set. Let E ∈ B. Then μ(E ∩ P_{1}) = μ(E ∩ P_{2}) and μ(E ∩ N_{1}) = μ(E ∩ N_{2}).

We keep our setup (X, B, μ). This lemma allows us to define two measures μ^{+} and μ^{−} in the following way: Let X = P ∪ N be a Hahn decomposition for X. Let E ∈ B. Then

μ^{+}(E) : = μ(E ∩ P)
μ^{−}(E) : = −μ(E ∩ N) .

Then μ^{+} and μ^{−} are measures and μ = μ^{+} − μ^{−}. We call μ^{+} the positive part of μ and μ^{−} the negative part of μ. The measure |μ| := μ^{+} + μ^{−} is called the total variation of μ.

Definition 7. Let μ and λ be measure on the same sigma algebra B. Then μ and λ are mutually singular (denoted by μ ⊥ λ) if there exists two disjoint sets A, B ∈ B such that X = A ∪ B and

μ(A) = λ(B) = 0.

We say that μ is absolutely continuous with respect to λ (denoted by μ << λ) if λ(E) = 0 implies μ(E) = 0.

Remark 2. We can also use the above phrases for signed measure. If μ and λ are signed measure then μ ⊥ λ if |μ| ⊥ |λ| and μ << λ if |μ| << |λ|.

Notice that μ(E) = μ(E ∩ A) and λ(E) = λ(E ∩ B) for all E ∈ B. Thus μ and λ are supported by disjoint sets. The consequence of the above discussion is the following theorem:

Theorem 11 (The Jordan Decomposition). Let μ be a signed measure defined on B. Then there are two mutually singular measure μ^{+} and μ^{−} on B such that

μ = μ^{+} − μ^{−} .

Moreover there is only one such set of mutually singular measures.

4.2. The Radon-Nikodym Theorem. Our first application of the Hahn decomposition if the proof of the Radon-Nikodym theorem, i.e., that μ << λ if and only if μ = λ_{f} for some f ∈ M^{+}(X).

Theorem 12 (Radon-Nikodym). Let μ and λ be measures defined on (X, B). Assume that X is σ−finite with respect to λ and that μ << λ. Then there exists a f ∈ M^{+}(X) such that μ = λ_{f}. The function f is unique up to set of measure zero.

Proof. Assume first that λ(X) < ∞. For c > 0 let • P (c): The positive set for the signed measure μ − cλ;

• N(c): The negative set for the signed measure μ − cλ.

Let

• A_{1} := N(c);

• A_{k+1} := N((k + 1)c) \ ⋃_{j=1}^{k} A_{j}.

Then

A_{k} = N(kc) ∩ ⋂_{j=1}^{k−1} P (jc) .

Let E ⊂ A_{k} be measurable. Then • (μ − kcλ)(E) ≤ 0 or μ(E) ≤ kcλ(E);

• For j < k : (μ − jcλ)(E) ≥ 0 or μ(E) ≥ jcλ(E).

In particular by taking j = k − 1 we get:

(k − 1)cλ(E) ≤ μ(E) ≤ kμ(E) .

Define now

A := ⋃_{j=1}^{∞} A_{j}  and  B = X \ A

Then μ = μ_{A} + μ_{B} and the same holds for λ. Let E ∈ B be contained in B. Then E is a positive set for μ − kcλ for all k. Thus μ(E) ≥ kcλ(E) for all k. Thus if λ(E) ≠ 0 then μ(E) = ∞. On the other hand if λ(E) = 0 then μ(E) = 0 because μ << λ. Hence we can define f(x) = ∞ for all x ∈ B.

Let us now work on A. Notice that A and B do not depend on the choice of c > 0! We will write A(c) and A_{k}(c) if we need the dependence of those sets on c. Take c = 2^{−n}, n ∈ ℕ. Notice that in this case we can choose the sets A and A_{k} such that

A_{k}(2^{−n}) = A_{2k−1}(2^{−(n+1)}) ∪ A_{2k}(2^{−(n+1)})

Define

f_{n}(x) := (k − 1)/2^{n} , x ∈ A_{k} .

Then the above shows that the sequence {f_{n}} is monotonic increasing. Define the function f on A by f = sup f_{n}. Then be the monotone convergence theorem it follows that

∫_{E} f dλ = sup ∫_{E} f_{n} dλ

for all E ∈ B, E ⊂ A. We also have

∫_{E∩A_{k}} f_{n} dλ = (k − 1)2^{−n}λ(E ∩ A_{k})
≤ μ(E ∩ A_{k})
≤ k2^{−1}λ(E ∩ A_{k})
= ∫_{E∩A_{k}} f_{n} dλ + 2^{−n}λ(E ∩ A_{k}) .

Thus

∫_{E∩A_{k}} f_{n} dλ ≤ μ(E ∩ A_{k}) ≤ ∫_{E∩A_{k}} f_{n} dλ + 2^{−n}λ(E ∩ A_{k}) .

Let ϵ > 0 be given. Choose N such that 2^{−N}λ(E ∩ A_{k}) < ϵ . (Notice that we are using the A_{k} defined by c = 2^{−N} so this set is fixed.) Then

∫_{E∩A_{k}} f_{n} dλ ≤ μ(E ∩ A_{k}) ≤ ∫_{E∩A_{k}} f_{n} dλ + ϵ ∀n ≥ N.

Thus

∫_{E∩A_{k}} fdλ ≤ μ(E ∩ A_{k}) ≤ ∫_{E∩A_{k}} f dλ + ϵ

As ϵ was arbitrary we finally get

μ(E ∩ A_{k}) = ∫_{E∩A_{k}} f dλ

This implies:

μ(E ∩ A) = ∑ μ(E ∩ A_{k})
= ∑ ∫_{E∩A_{k}} f dλ
= ∫_{E∩A} f dλ

This shows that μ = λ_{f}.

If λ(X) = ∞, then we choose a disjoint sequence {X_{n}} such that X = ∪X_{n} and μ(X_{n}) < ∞ for all n. Let f_{n} be the function constructed above for λ and μ restricted to X_{n}. We view f_{n} as a function on X be letting f|_{X\X_{n}} = 0. Let f = ∑ f_{n}. Then μ = λ_{f}.

Theorem 13 (Lebesgue Decomposition). Let (X, B, μ) be a σ-finite measure space and let ν be a σ-finite measure defined on B. Then we can find measures ν_{0} and ν_{1} such that:

1. ν_{0} is singular with respect to μ;

2. ν_{1} << μ;

3. ν = ν_{0} + ν_{1}.

The measures ν_{0} and ν_{1} are unique.

Proof. Let λ := ν + μ. Then λ is σ-finite and ν, μ << λ. Choose f, g ∈ M^{+} such that

μ(E) = ∫_{E} f dλ and ν(E) = ∫_{E} g dλ .

Let A := {x | f(x) > 0} and B := {x | f(x) = 0}. Then A, B ∈ B and X = A ˙∪B. Define ν_{0} and ν_{1} by

ν_{0}(E) := ν(E ∩ B) and ν_{1}(E) := ν(E ∩ B) .

Then ν = ν_{0} + ν_{1}. By construction we obviously have ν_{0} ⊥ μ. On the other hand assume that μ(E) = 0. Then f(x) = 0 for λ-almost all x ∈ E. We also have E = (E ∩ A) ∪ (E ∩ B). Hence λ(E ∩ B) = 0. It follows that

ν_{0}(E) = ν(E ∩ B)
= ∫_{E∩B} g dλ
= 0

Hence ν_{0} << μ. The uniqueness is obvious and is left to the reader.

4.3. Riesz Representation Theorem. The next application is the proof of the Riesz Representation Theorem:

Theorem 14 (The Riesz Representation Theorem). Let (X, B, μ) be a σ-finite measure space. Let 1 ≤ p < ∞ and let F ∈ L^{p}(X)^{∗}. Then there exists a g ∈ L^{q}(X), where 1/p + 1/q = 1, such that

< f, F >= ∫ fg dμ ∀f ∈ L^{p}(X) .

We need two lemmas before we start the proof and the space L^{∞}(X).

Lemma 15. Let 1 ≤ p < ∞ and let q be given by 1/p + 1/q = 1. Let (X, B, μ) be a finite measure space and g ∈ L^{1}(X) such that there exists a constant M > 0 with

|∫ gφ dμ| ≤ M ||φ||_{p}

for all simple functions φ. Then g ∈ L^{q}(X).

Proof. See Book p. 283.

Lemma 16. Let {E_{n}} be a sequence of disjoint measurable sets, and for each n let f_{n} be a function in L^{p}(X), 1 ≤ p < ∞, that vanishes outside E_{n}. Let f := ∑ f_{n}. Then f ∈ L^{q}(X) if and only if ∑ ||f_{n}||_{p}^{p} < ∞. In this case f = ∑ f_{n} in L^{p}(X) and

||f||_{p} = \sqrt[p]{∑ ||f_{n}||_{p}^{p}} .

Proof. We can assume that X = ∪E_{n}. Let x ∈ X then there is an unique n such that x ∈ E_{n}. Hence f(x) = f_{n}(x). In particular |f(x)|^{p} = |f_{n}(x)|^{p}. It follows (using the monotone convergence theorem) that

∫ |f(x)|^{p} dμ(x) = ∑ ∫_{E_{n}} |f(x)|^{p} dμ(x)
= ∑ ∫_{E_{n}} |f_{n}(x)|^{p} dμ(x)
= ∑ ||f_{n}||^{p}
< ∞

In the same way

||f − ∑_{j=1}^{N} f_{n}||_{p}^{p} = || ∑_{j=N+1}^{∞} f_{n}||_{p}^{p}

≤ ∑_{j=N+1}^{∞} ||f_{n}||_{p}^{p}

and ∑_{j=N+1}^{∞} ||f_{n}||_{p}^{p} can be made arbitrary small. From this the Lemma follows.

We can now proof the Riesz Representation Theorem:

Proof. (Of the Riesz Representation Theorem) We can obviously assume that μ(X) > 0. Assume first that μ(X) < ∞. Define

ν(E) :=< F, χ_{E} > , E ∈ B .

Then |ν(E)| ≤ ||F || ||χ_{E}|| ≤ ||F || μ(X)^{1/p} for all E ∈ B. Moreover:

1. ν(∅) = 0 because χ_{∅} = 0.

2. Let E = ⋃_{n=1}^{∞} E_{n} be a disjoint union. Then

χ_{E} = ∑_{n=1}^{∞} χ_{E_{n}} (in L^{p})

Let f = ∑(sign(ν(E_{n}))χ_{En}. Then f ∈ L^{p} and

< F, f >= ∑_{n=1}^{∞} |ν(E_{n})| .

Hence ∑_{n=1}^{∞} |ν(E_{n})| converges absolutely and it follows that

ν(E) = ∑_{n=1}^{∞} ν(E_{n}) .

Hence ν is a signed measure. Let ν = ν^{+} − ν^{−}. Suppose that μ(E) = 0. Then χ_{E} = 0 almost everywhere. Let P (respectively N) be the positive (respectively negative) set for ν. Then μ(E ∩ P) = μ(E ∩ N) = 0. In particular < F, χ_{E∩P} >=< F, χ_{E∩N} >= 0. Thus

ν^{+}(E) = ν(E ∩ P) =< F, χ_{E∩P} >= 0

and similarly ν^{−}(E) = 0. It follows that ν^{+} and ν^{−} are absolutely continuous with respect to μ. Thus there are g^{+}, g^{−} ∈ M^{+} such that

ν^{+} = μ_{g^{+}} and ν^{−} = μ_{g^{−}} .

Hence ν = μ_{g} where g = g^{+} − g^{−}. As ν^{+}(X), ν^{−}(X) < ∞ it follows that g ∈ L^{1}. We need to show that g ∈ L^{q}(X). Let

f(x) = \frac{sign(g(x))}{μ(X)^{1/p}} |g(x)|^{p−1} .

5. Measure and Outer Measure

5.1. Outer Measures and Measurability.

5.2. The Extension Theorem.

Definition 8. A collection A of subsets of X is called an algebra of sets if 1. If A, B ∈ A then A ∪ B ∈ A;

2. If A ∈ A then A^{c} ∈ A.

Lemma 17. Let A be an algebra of sets. Let E_{j} ∈ A, j = 1, …, n be a finite collection of sets in A. Then ∪_{j=1}^{n}E_{j} , ∩_{j=1}^{n}E_{j} ∈ A.

Proof. It suffice to show that with A, B ∈ A it follows that A∩B ∈ A. The Lemma follows then by induction. But A ∩ B = (X \ [A^{c} ∪ B^{c}]) ∈ A.

Definition 9. A measure μ on an algebra A is a function μ : A → [0, ∞] such that 1. μ(∅) = 0;

2. If {E_{i}} is a disjoint sequence of sets in A such that ∪E_{i} ∈ A. Then

μ(∪E_{i}) = ∑ μ(E_{i}) .

Thus μ is a measure if and only if A is a σ-algebra. We will show that if μ is a measure on a algebra A then there exists a σ-algebra B containing A and a measure μ̄ on B that extends μ, i.e., μ̄(E) = μ(E) for all E ∈ A. We do this by first constructing an outer measure μ^{∗} that extends μ. Then the extension theorem from the last section guaranties the existence of μ̄. We define μ^{∗} by

μ^{∗}(B) = inf ∑_{j=1}^{∞} μ(A_{i})

where the inf is taking over all sequences {A_{i}} in A such that B ⊂ ∪A_{i}.

Lemma 18. The function μ^{∗} : P(X) → ℝ̄ satisfies the following:

1. μ^{∗}(∅) = 0;

2. μ^{∗}(B) ≥ 0;

3. If A ⊂ B then μ^{∗}(A) ≤ μ^{∗}(B);

4. If A ∈ A then μ^{∗}(A) = μ(A);

5. If {A_{n}} is a sequence of sets then

μ^{∗}(∪A_{i}) ≤ ∑ μ^{∗}(A_{i}) .

In particular it follows that μ^{∗} is an outer measure.

Proof. (1)-(3) follows directly from the definition.

4) As A ⊂ A ∪ ∅ ∪ ∅ ⋯ if follows that μ(A) ≥ μ^{∗}(A). On the other hand if A ⊂ ∪E_{n} then A = ∪(A ∩ E_{n}) = ∪(A ∩ F_{n}) where F_{1} := E_{1}, and

F_{n+1} = E_{n+1} \ ∪_{j=1}^{n}F_{j} = E_{n+1} \ ∪_{j=1}^{n}E_{j}

Hence

μ(A) ≤ ∑ μ(A ∩ F_{n}) ≤ ∑ μ(A ∩ E_{n}) ≤ ∑ μ(E_{n}).

Hence μ(A) ≤ μ^{∗}(A).

5) Let ϵ > 0. For each n choose {E_{n,k}} such that

A_{n} ⊂ ∪E_{n,k}

and

∑_{k} μ(E_{n,k}) ≤ μ^{∗}(A_{n}) + ϵ2^{−n} .

Then

∪A_{n} ⊂ ∪_{n} ∪_{k} E_{n,k}

and hence

μ^{∗}(∪A_{n}) ≤ ∑_{n,k} μ(E_{n,k})
≤ ∑_{n} μ^{∗}(A_{n}) + ϵ

Since ϵ was arbitrary it follows that μ (∪A_{n}) ≤ ∑ μ^{∗}(A_{n}).

Lemma 19. Let B be the σ-algebra given by μ^{∗}. Then A ⊂ B.

Proof. Let E ∈ A and let A ⊂ X. Let ϵ > 0 be given. Let {A_{n}} be a sequence in A such that

A ⊂ ⋃_{n=1}^{∞} A_{n}

and

∑ μ(A_{n}) ≤ μ^{∗}(A) + ϵ .

Then, since A∩E ⊂ ∪(A_{n} ∩E) and A∩E^{c} ⊂ ∪(A_{n} ∩E^{c}) and A_{n} ∩E_{n}, A_{n} ∩E_{n}^{c} ∈ A it follows that

μ^{∗}(A ∩ E) ≤ ∑ μ(A_{n} ∩ E)

and

μ^{∗}(A ∩ E^{c}) ≤ ∑ μ(A_{n} ∩ E^{c}) .

Thus

μ^{∗}(A) ≤ μ^{∗}(A ∩ E) + μ^{∗}(A ∩ E^{c})
≤ ∑ μ(E ∩ A_{n}) + ∑ μ(E^{c} ∩ A_{n})
= ∑ μ(A_{n})
≤ μ^{∗}(A) + ϵ

As ϵ > 0 was arbitrary it follows that μ^{∗}(A) = μ^{∗}(A ∩ E) + μ^{∗}(A ∩ E^{c}). Hence E is measurable.

By the extension theorem we get now:

Theorem 15 (Carathéodory Extension Theorem). Let μ be a measure on the algebra A. Then there exists a σ-algebra B containing A and a complete measure μ̄ on B such that μ̄(A) = μ(A) for all A ∈ A. If μ is finite then so is μ̄.

Theorem 16 (Hahn Extension Theorem). Let μ be a σ-finite measure on the algebra A. Let B be the σ-algebra of μ^{∗}-measurable sets. Then the extension μ̄ of μ is unique and μ̄ is σ-finite.

Proof. Suppose first that μ is finite. Let ν be a measure on B that agrees with μ on A. We notice that both μ̄ and ν are finite. Let E ∈ B. Let {E_{n}} be a sequence in A such that E ⊂ ∪E_{n}. Then

ν(E) ≤ ν(∪E_{n})
≤ ∑ ν(E_{n})
= ∑ ν(E_{n}) .

Hence

ν(E) ≤ μ̄(E) .

But we also have

∞ > μ(X) = μ̄(E) + μ̄(E^{c})
= ν(E) + ν(E^{c})

because X = E ˙∪E^{c} and both μ̄ and ν are additive. But by the first part we have

ν(E) ≤ μ̄(E) and ν(E^{c}) ≤ μ̄(E^{c})

Hence ν(E) = μ̄(E) and ν(E^{c}) = μ̄(E^{c}). This complete the proof for μ(X) < ∞.

Assume now that μ is σ-finite. Choose X_{n} such that μ(X_{n}) < ∞ and X = ∪X_{n}. We may assume that the sets X_{n} are disjunct. By the first part it follows that μ̄|X_{n} = ν|X_{n}. Let E ∈ B. Then E = ∪(X_{n} ∩ E). Hence

μ̄(E) = ∑ μ̄(E ∩ X_{n})
= ∑ ν(E ∩ X_{n})
= ν(E)

Hence the claim.

Let A be an algebra. Then P(X) is a σ-algebra containing A. Let

A^{∗} = ⋂_{A⊂C σ-algebra} C

Then A^{∗} is the smallest σ-algebra containing A. We call A^{∗} the σ-algebra generated by A. We notice that B from the above theorem is a σ-algebra containing A. Hence A^{∗} ⊂ B, but B can be much bigger. But the Theorem guaranties that we can extend μ to a measure on A^{∗}, by restricting μ̄ to A^{∗}.

5.3. Product Measures. Our first application is the construction of a product measure and Fubini’s Theorem. Let (X, A, μ) and (Y, B.ν) be measure spaces. Let

Z := X × Y = {(x, y) | x ∈ X, y ∈ Y }

We call the sets

A × B , A ∈ A , B ∈ B

for measurable rectangle. Denote the collection of all finite union of measurable rectangles by C_{0}.

Lemma 20. Let E ∈ C_{0}. Then E is a finite collection of disjoint measurable rectangles.

Proof. By induction we can reduce the proof to the situation where

E = A × B ∪ C × D .

But then

E = (A ∩ C) × (B ∩ D) ˙∪(A \ C) × B ˙∪(A ∩ C) × (B \ D) ˙∪(A ∩ C) × (D \ B) ˙∪(C \ A) × D .

Lemma 21. Let the notation be as above. Then C_{0} is an algebra of sets.

Proof. We have by definition that any finite union of sets in C_{0} is again in C_{0}. Let A ∈ A and B ∈ B. Then

(A × B)^{c} = {(x, y) | x ∉ A or y ∉ B}
= A^{c} × Y ∪ X × B^{c}

We denote by A × B the σ-algebra generated by C_{0}. Define a function on C_{0} by

λ( ˙∪_{i=1}^{n}A_{i} × B_{i}) = ∑_{i=1}^{n} μ(A_{i})ν(B_{i})

We have to show that λ is well defined. Then λ is a measure on the algebra C_{0}. Denote by μ × ν the measure on A × B generated by λ.

Theorem 17 (Product Measure). Let (X, A, μ) and (Y, B, ν) be measure spaces. Then there is a complete measure μ × ν on A × B such that

μ × ν(A × B) = μ(A)ν(B) ∀A ∈ A, ∀B ∈ B . (5.1)

If μ and ν are both σ-finite, then μ × ν is σ-finite and the measure μ × ν is uniquely determined by 5.1. The measure μ × ν is called the product of μ and ν.

Proof. Define λ as above. Let

E = ⋃_{i=1}^{n} A_{i} × B_{i} = ⋃_{j=1}^{k} C_{j} × D_{j}

be a disjoint union of rectangles. Then

χ_{E}(x, y) = ∑ χ_{A_{i}}(x)χ_{B_{i}}(y)
= ∑ χ_{C_{j}}(x)χ_{D_{j}}(y)

Keep x fixed and integrate over Y given (recall that we are dealing with finite sums)

∑ χ_{A_{i}}(x)ν(B_{i}) = ∑ χ_{C_{j}}(x)ν(D_{j}) .

Integrating over X gives then

∑ μ(A_{i})ν(B_{i}) = ∑ μ(C_{j})ν(D_{j}) .

Hence λ(E) is well defined.

The properties λ(∅) = 0 and λ(A) ≤ λ(B) if A ⊂ B are obvious. Thus we only have to show that λ is σ-additive. Let {E_{i}} ∈ C_{0} be disjoint and such that ∪_{i}E_{i} ∈ C_{0}. We have to show that

λ (⋃ E_{i}) = ∑ λ(E_{i}) .

As the union as well as each E_{i} can be written as finite union of disjoint measurable rectangle, we can assume that E = A × B = ⋃_{i=1}^{∞} A_{i} × B_{i}, disjoint union. Then again

χ_{E}(x, y) = χ_{A}(x)χ_{B}(y) = ∑_{i=1}^{∞} χ_{A_{i}}(x)χ_{B_{i}}(y) .

Integrating and using Theorem 4 and the fact that all the functions and number occurring are nonnegative, we get

χ_{A}(x)ν(B) = ∑_{i=1}^{∞} χ_{A_{i}}(x)ν(B_{i})

Integrating over X and using Theorem 4 again gives

μ(A)ν(B) = ∑_{i=1}^{∞} μ(A_{i})ν(B_{i}) .

To generalize this to finite union is now clear.

Assume that μ and ν are σ-finite. Let X = ∪X_{n} and Y = ∪Y_{m} with μ(X_{n}) < ∞ and ν(Y_{n}) < ∞. Then

X × Y = ⋃_{n,m} X_{n} × Y_{m}

and λ(X_{n} × Y_{m}) = μ(X_{n})ν(Y_{m}) < ∞.

The theorem follows now from the Carathéodory Extension Theorem and Hahn Extension Theorem.

We can now integrate functions on X × Y with respect to the measure μ × ν. We would like to show that the integral can be evaluated by integrating first the one variable and then the other, i.e.,

∫_{X×Y} h(x, y) d (μ × ν) (x, y) = ∫_{X} (∫_{Y} h(x, y) dν(y)) dμ(x)
= ∫_{X} (∫_{Y} h(x, y) dμ(x)) dν(y) .

As usually the strategy is to prove this first for the characteristic function χ_{E}. That implies the claim for simple functions. We can then proof it for positive functions h by choosing an increasing sequence of simple functions, φ_{n} ↗ h. The general case follows then by applying this to h^{+} and h^{−}.

Let E ⊂ X ×Y. Let x ∈ X and y ∈ Y. We define the x-section and the y-section of E to be the sets:

E_{x} : = {y ∈ Y | (x, y) ∈ E} ;
E^{y} : = {x ∈ X | (x, y) ∈ E} .

Similarly if f is a function defined on X × Y we define the x-section and the y-section of f to be the functions:

f_{x}(y) := f(x, y)  and  f^{y}(x) := f(x, y) .

Notice that (χ_{E})_{x} = χ_{E_{x}} and similarly for y. We remark the following simple fact:

Lemma 22. Let the notation be as above. Then the following holds:

1. (E_{x})^{c} = (E^{c})_{x} and (E^{y})^{c} = (E^{c})^{y}.

2. (∪_{i}E_{i})_{x} = ∪ (E_{i})_{x} and (∪E_{i})^{y} = ∪ (E_{i})^{y}.

Proof. We will only show the statement for x as the proof for y is exactly the same. 1) We have

y ∈ (E_{x})^{c} ⇔ y ∉ E_{x} ⇔ (x, y) ∉ E ⇔ (x, y) ∈ E^{c} ⇔ y ∈ (E^{c})_{x} .

2) Similarly we have

y ∈ (∪E_{i})_{x} ⇔ (x, y) ∈ ∪E_{i}
⇔ ∃i : (x, y) ∈ E_{i}
⇔ ∃i : y ∈ (E_{i})_{x}
⇔ y ∈ ∪E_{ix}

Definition 10. A class M of sets is called monotone if it is non-empty and M contains all union of increasing sequences of elements in M as well as the intersection of all decreasing sequences of elements in M.

Lemma 23. Let C be an algebra of sets. Let A be the σ-algebra generated by C and let B be the smallest algebra monotone class containing C. Then A = B.

Proof. Obviously B ⊂ A. To show the opposite we need to show that B is an algebra, because then each countable union can be written as union of increasing sets from B.

Let E ∈ B. Define

B(E) := {F ∈ B |E \ F, F \ E, E ∩ F ∈ B} .

Then E, ∅ ∈ B(E) and B(E) is a monotone class. Moreover F ∈ B(E) if and only if E ∈ B(F).

Suppose that E ∈ C. Then - because C is an algebra- we have C ⊂ B(E). Hence - as B was the smallest monotone class generated by C - we get B(E) = B.

Let now F ∈ B be arbitrary. Then F ∈ B(E). Hence E ∈ B(F). It follows that C ⊂ B(F). So again we get B(F) = B. Thus B is closed under taking complement and finite intersection. Hence it is also closed under taking finite union. Hence it is an algebra and then also a σ-algebra.

Lemma 24. Let the notation be as above. Let x ∈ X and y ∈ Y. Then the following holds:

1. If E ∈ A × B then E_{x} ∈ B and E^{y} ∈ A.

2. If f is measurable then f_{x} and f^{y} are measurable.

Proof. 1) Assume first that E = A×B. Then E_{x} = ∅ ∈ B if x ∉ A and E_{x} = B ∈ B if x ∈ A. Also if

E = ⋃_{i=1}^{n} A_{i} × B_{i} ∈ C_{0}

is a disjoint union of measurable rectangles, then

E_{x} = ⋃_{i, x∈A_{i}} B_{i} ∈ C_{0} .

Define E = {E ∈ A × B | E_{x} ∈ B}. Then C_{0} ⊂ E. Let {E_{n}} be an increasing sequence in E. Then

(∪E_{n})_{x} = ∪E_{nx} ∈ B .

Similarly we show that the intersection of a decreasing sequence is back in E. It follows that E is a monotone class containing C_{0}. Hence E = A × B.

The claim for E^{y} is shown in the same way.

2) We have for α ∈ ℝ:

{y ∈ Y | f_{x}(y) > α} = {y ∈ Y | f(x, y) > α}
= {(z, y) | f(z, y) > α}_{x} ∈ B .

The claim for f^{y} follows in the same way.

Lemma 25. Let (X, A, μ) and (Y, B, ν) be σ-finite measure spaces. Let E ∈ A × B, x ∈ X and y ∈ Y. Define

f(x) = ν(E_{x})

and

g(y) = μ(E^{y}) .

Then f and g are measurable and

∫ f(x) dμ(x) = ∫ g(y) dν(y) = μ × ν(E) .

Proof. Assume first that X and Y are finite. Let

E := {E ∈ A × B | f is measureable and ∫ f = μ × ν(E) }

We will show that E = A × B. We do that in few steps.

1) If E = A × B then f(x) = χ_{A}(x)ν(B). Hence E ∈ E. In particular it follows that E ≠ ∅.

2 Let {E_{n}} be an increasing sequence in E then ∪E_{n} ∈ E: So let {E_{n}}_{n} be a sequence with E_{n} ⊂ E_{n+1}. Then

( ⋃_{j=1}^{N} E_{i})_{x} = (E_{N})_{x}

so (⋃_{j=1}^{N} E_{i}) ∈ E. Let f_{N}(x) = ν(E_{Nx}) and f(x) = ν((∪E_{n})_{x}). Then {f_{N}} is an increasing sequence with

f(x) = lim_{N→∞} f_{N}(x)

so f is measurable. By the monotone convergence theorem we get

∫ f dμ = lim ∫ f_{N} dμ = lim μ × ν(E_{N}) = μ × ν(∪E_{N})

by Problem p. 257. Hence ∪E_{n} ∈ E.

3) Similarly we show that if {E_{n}} is a decreasing sequence in E then ∩E_{n} ∈ E. Thus E is a monotone class containing the algebra C_{0}. Hence E = A × B. The theorem now follows in the usual way writing X × Y = ∪X_{n} × Y_{n} with X_{n}

and Y_{n} finite and by using the monotone convergence theorem.

Theorem 18 (Tornelli’s Theorem). Let (X, A, μ) and (Y, B, ν) be σ-finite measure spaces and let h : X × Y → [0, ∞] be measurable. Then the functions

f(x) := ∫ h_{x}(y) dν(y) = ∫ h(x, y) dν(y)

and

g(y) := ∫ h^{y}(x) dμ(x) = ∫ h(x, y) dμ(x)

are measurable and

∫ f(x) dμ(x) = ∫ g(y) dν(y) = ∫ h(x, y) dμ × ν(x, y)

Proof. Suppose first that h = χ_{E}. Then h(x, y) = χ_{E_{x}}(y) and

∫_{Y} h(x, y) dν(y) = ∫_{Y} χ_{E_{x}}(y) dν(y)
= ν(E_{x}) .

By the last lemma it follows that x ↦ ν(E_{x}) is measurable and

∫_{X} (∫_{Y} h(x, y) dν(y)) dμ(x) = ∫_{X} ν(E_{x}) dμ(x) = μ × ν(E) = ∫_{X×Y} h d(μ × ν) .

Similarly it follows that

∫_{Y} ∫_{X} h(x, y) dμ(x)dν(y) = ∫_{X×Y} h(x, y) d(μ × ν)(x, y) .

From this it follows that the claim holds for simple functions. Let now h ∈ M^{+}. Choose an increasing sequence {φ_{n}} of simple function approximating h. Then (by the monotone convergence theorem)

∫_{X×Y} h = lim ∫_{X×Y} φ_{n}
= lim ∫_{X} (∫_{Y} φ_{n}(x, y) dν(y)) dμ(x)
= lim ∫_{Y} (∫_{X} φ_{n}(x, y) dμ(x)) dν(x) .

As φ_{n} ↗ h it follows that ∫ φ_{n}(x, y) dν(y) is an increasing sequence. Hence

lim ∫_{X} (∫_{Y} φ_{n}(x, y) dν(y)) dμ(x) = ∫_{X} lim (∫_{Y} φ_{n}(x, y) dν(y)) dμ(x)

But φ_{nx} ↗ h_{x}. Hence by the monotone convergence theorem we have

lim ∫_{Y} φ_{n}(x, y) dν(y) = ∫_{Y} lim φ_{n}(x, y) dν(y)
= ∫_{Y} h(x, y) dν(y) .

As each of the functions x ↦ ∫_{Y} φ_{n}(x, y) dν(y) are measurable it follows that x ↦ ∫_{Y} h(x, y) dν(y) is measurable. Furthermore

∫_{X×Y} h = lim ∫_{X} (∫_{Y} φ_{n}(x, y) dν(y)) dμ(x)
= ∫_{X} (lim ∫_{Y} φ_{n}(x, y) dν(y)) dμ(x)
= ∫_{X} (∫_{Y} lim φ_{n}(x, y) dν(y)) dμ(x)
= ∫_{X} (∫_{Y} h(x, y) dν(y)) dμ(x) .

We show in the same way that ∫_{X×Y} h = ∫_{Y} ∫_{X} h(x, y)dμ(x)dν(y). From that the theorem follows.

Theorem 19 (Fubini’s Theorem). Let (X, A, μ) and (Y, B, ν) be σ-finite measure spaces and let h ∈ L^{1}(X ×Y, μ×ν). Then the extended real-valued functions defined almost everywhere by

f(x) := ∫ h_{x}(y) dν(y) = ∫ h(x, y) dν(y)

and

g(y) := ∫ h^{y}(x) dμ(x) = ∫ h(x, y) dμ(x)

are integrable and

∫ f(x) dμ(x) = ∫ g(y) dν(y) = ∫ h(x, y) dμ × ν(x, y)

Proof. We use Tonelli’s Theorem for the functions h^{+} and h^{−}.

We remark that it is necessary to assume that the function h is measurable to make sure that the integrals involving the functions h_{x}^{+}, h^{+y}, h_{x}^{−} and h^{−y} are finite for almost all y and x. But Tonelli’s theorem can be used to actually proof that the function h is integrable because

∫_{X×Y} |h(x, y)| = ∫_{X} (∫_{Y} |h(x, y)| dν(y)) dμ(x)
= ∫_{Y} (∫_{X} |h(x, y)| dμ(x)) dν(y) .

Hence h is integrable if and only if the integral

∫_{Y} |h(x, y)| dν(y)

30 GESTUR ÓLAFSSON

is integrable as a function of x (so in particular it has to be finite for almost all x ∈ X).

Department of Mathematics, LSU, Baton Rouge, LA 7083

E-mail address: olafsson@math.lsu.edu