Integration

Measurable Partition

Suppose that

𝒮

is a

σ

-algebra on a set

X

. An $𝒮$ -partition of

X

is a finite collection

A_{1}, \dots, A_{m}

of pairwise disjoint sets in

𝒮

such that

A_{1} \cup \dots \cup A_{m} = X .

Lower Lebesgue Sum

Suppose that

(X, 𝒮, μ)

is a measure space,

f : X \to [0, \infty]

𝒮

-measurable, and

P = (A_{1}, \dots, A_{m})

is an

𝒮

-partition of

X

. The lower Lebesgue sum of

f

over

P

with respect to

μ

L (f, P, μ) : = \sum_{j = 1}^{m} μ (A_{j}) \inf_{A_{j}} f .

Integral of a Nonnegative Measurable Function

Suppose that

(X, 𝒮, μ)

is a measure space and

f : X \to [0, \infty]

𝒮

-measurable. The integral of

f

with respect to

μ

\int f d μ : = \sup {L (f, P, μ) : P is an 𝒮 -partition of X} .

Integral of a Characteristic Function

Suppose that

(X, 𝒮, μ)

is a measure space and

E \in 𝒮

. Then

\int χ_{E} d μ = μ (E) .

For any $𝒮$ -partition $P$ , the lower sum for $χ_{E}$ counts only the parts contained in $E$ , so $L (χ_{E}, P, μ) \leq μ (E)$ . Taking the partition $E, X ∖ E$ gives $L (χ_{E}, P, μ) = μ (E)$ . Hence the supremum in the definition of the integral is $μ (E)$ .

Integral of a Simple Function

Suppose that

(X, 𝒮, μ)

is a measure space,

E_{1}, \dots, E_{n} \in 𝒮

are pairwise disjoint, and

c_{1}, \dots, c_{n} \in [0, \infty]

. Then

\int (\sum_{k = 1}^{n} c_{k} χ_{E_{k}}) d μ = \sum_{k = 1}^{n} c_{k} μ (E_{k}) .

Refine any partition by the measurable sets $E_{1}, \dots, E_{n}$ . On each $E_{k}$ , the function has constant value $c_{k}$ , and outside their union it has value $0$ . Applying the characteristic-function computation and finite additivity of measure gives $\int (\sum_{k = 1}^{n} c_{k} χ_{E_{k}}) d μ = \sum_{k = 1}^{n} c_{k} μ (E_{k}) .$

Integration of Nonnegative Functions is Order Preserving

Suppose that

(X, 𝒮, μ)

is a measure space and

f, g : X \to [0, \infty]

are

𝒮

-measurable. If

f (x) \leq g (x)

for every

x \in X

, then

\int f d μ \leq \int g d μ .

If $P$ is an $𝒮$ -partition, then $\inf_{A} f \leq \inf_{A} g$ for every part $A \in P$ . Thus $L (f, P, μ) \leq L (g, P, μ)$ . Taking suprema over all partitions gives $\int f d μ \leq \int g d μ$ .

Monotone Convergence Theorem

Suppose that

(X, 𝒮, μ)

is a measure space and

0 \leq f_{1} \leq f_{2} \leq \dots

is an increasing sequence of

𝒮

-measurable functions. Define

f : X \to [0, \infty]

f (x) : = \lim_{k \to \infty} f_{k} (x) .

Then

\lim_{k \to \infty} \int f_{k} d μ = \int f d μ .

By order preservation, $\int f_{k} d μ \leq \int f d μ$ , so the limit of the integrals is at most $\int f d μ$ . For the reverse inequality, take any lower sum $L (f, P, μ)$ and replace each infimum on a part by a slightly smaller value. Since $f_{k} ↑ f$ , the sets on which $f_{k}$ exceeds those smaller values increase to each part. The measure of an increasing union then shows that ${lim inf}_{k} \int f_{k} d μ \geq L (f, P, μ)$ . Taking the supremum over $P$ proves equality.

Additivity of Integration for Nonnegative Functions

Suppose that

(X, 𝒮, μ)

is a measure space and

f, g : X \to [0, \infty]

are

𝒮

-measurable. Then

\int (f + g) d μ = \int f d μ + \int g d μ .

First prove the result when $f$ and $g$ are nonnegative simple functions by refining their measurable partitions and using the formula for the integral of a simple function. For general nonnegative measurable $f$ and $g$ , choose increasing simple approximations from approximation by simple functions. Apply the simple case to the approximants and pass to the limit using the monotone convergence theorem.

Positive Part of a Function

Suppose that

f : X \to [- \infty, \infty]

. The positive part of

f

is the function

f^{+} : X \to [0, \infty]

defined by

f^{+} (x) : = {\begin{matrix} f (x) & f (x) \geq 0, \\ 0 & f (x) < 0. \end{matrix}

Negative Part of a Function

Suppose that

f : X \to [- \infty, \infty]

. The negative part of

f

is the function

f^{-} : X \to [0, \infty]

defined by

f^{-} (x) : = {\begin{matrix} 0 & f (x) \geq 0, \\ - f (x) & f (x) < 0. \end{matrix}

Integral of an Extended Real-Valued Function

Suppose that

(X, 𝒮, μ)

is a measure space and

f : X \to [- \infty, \infty]

𝒮

-measurable. Let

f^{+}

be the positive part of

f

, and let

f^{-}

be the negative part of

f

. If at least one of

\int f^{+} d μ

and

\int f^{-} d μ

is finite, define

\int f d μ : = \int f^{+} d μ - \int f^{-} d μ .

Additivity of Integration

Suppose that

(X, 𝒮, μ)

is a measure space and

f, g : X \to ℝ

are

𝒮

-measurable. If

\int | f | d μ < \infty and \int | g | d μ < \infty,

then

\int (f + g) d μ = \int f d μ + \int g d μ .

Write $f = f^{+} - f^{-}$ and $g = g^{+} - g^{-}$ , using the positive and negative parts. The hypotheses imply all four nonnegative integrals are finite. Since ${(f + g)}^{+} + f^{-} + g^{-} = {(f + g)}^{-} + f^{+} + g^{+},$ applying additivity for nonnegative functions to both sides and rearranging gives $\int (f + g) d μ = \int f d μ + \int g d μ$ .

Integration on a Subset

Suppose that

(X, 𝒮, μ)

is a measure space,

E \in 𝒮

, and

f : X \to [- \infty, \infty]

𝒮

-measurable. Define

\int_{E} f d μ : = \int χ_{E} f d μ

whenever the integral on the right is defined.

Bound Variable Notation for Integrals

Suppose that

(X, 𝒮, μ)

is a measure space,

E \in 𝒮

, and an expression

A (x)

defines a measurable function

g : X \to [- \infty, \infty], g (x) = A (x) .

Then

\int_{E} A (x) d μ (x) : = \int_{E} g d μ,

whenever the integral on the right is defined. The notation

d μ (x)

means that

x

is the bound variable of integration; it is not a new measure and it is not multiplication by

d x

Other symbols appearing in $A (x)$ are treated as fixed parameters unless they are also bound by an integral or quantifier. For example, if $(Ω, 𝒮, σ)$ is a measure space, $D \in 𝒮$ , and $x$ is fixed, then $\int_{D} | ω \cdot N (x) | d σ (ω)$ means $\int_{D} g_{x} d σ, g_{x} (ω) = | ω \cdot N (x) | .$ Here $ω$ is the variable being integrated over, while $x$ is just a parameter.

Almost Every

Suppose that

(X, 𝒮, μ)

is a measure space. A set

E \in 𝒮

contains $μ$ -almost every element of

X

μ (X ∖ E) = 0.

Bounded Convergence Theorem

Suppose that

(X, 𝒮, μ)

is a measure space with

μ (X) < \infty

. Suppose that

f_{1}, f_{2}, \dots : X \to ℝ

are

𝒮

-measurable and converge pointwise to

f : X \to ℝ

. If there exists

c \in ℝ^{+}

such that

| f_{k} (x) | \leq c

for every

k \in ℤ^{+}

and every

x \in X

, then

\lim_{k \to \infty} \int f_{k} d μ = \int f d μ .

Since $| f_{k} | \leq c$ and $μ (X) < \infty$ , the constant function $c$ is integrable. The functions $c + f_{k}$ and $c - f_{k}$ are nonnegative and converge pointwise to $c + f$ and $c - f$ . Applying the monotone convergence theorem to suitable increasing lower envelopes, or equivalently applying Fatou's lemma as derived from monotone convergence, gives $\int f d μ \leq \underset{k}{lim inf} \int f_{k} d μ \leq \underset{k}{lim sup} \int f_{k} d μ \leq \int f d μ .$ Therefore the integrals converge to $\int f d μ$ .

Dominated Convergence Theorem

Suppose that

(X, 𝒮, μ)

is a measure space,

f : X \to [- \infty, \infty]

𝒮

-measurable, and

f_{1}, f_{2}, \dots : X \to [- \infty, \infty]

are

𝒮

-measurable. Suppose that

f_{k} \to f

for almost every

x \in X

. If there exists an

𝒮

-measurable

g : X \to [0, \infty]

such that

\int g d μ < \infty and | f_{k} (x) | \leq g (x)

for every

k \in ℤ^{+}

and almost every

x \in X

, then

\lim_{k \to \infty} \int f_{k} d μ = \int f d μ .

After changing the functions on a null set, assume the pointwise convergence and domination hold everywhere. Since $g + f_{k} \geq 0$ and $g - f_{k} \geq 0$ , Fatou's lemma, obtained from the monotone convergence theorem, gives $\int g d μ + \int f d μ \leq \underset{k}{lim inf} (\int g d μ + \int f_{k} d μ)$ and the analogous inequality for $g - f_{k}$ . Because $\int g d μ < \infty$ , subtracting $\int g d μ$ yields matching lower and upper bounds for $\int f_{k} d μ$ , so the limit is $\int f d μ$ .

Riemann Integrable iff Continuous Almost Everywhere

Suppose that

a < b

and

f : [a, b] \to ℝ

is bounded. Then

f

is Riemann integrable if and only if

omes ({x \in [a, b] : f is not continuous at x}) = 0.

Moreover, if

f

is Riemann integrable and

λ

is Lebesgue measure on

ℝ

, then

f

is Lebesgue measurable and

\int_{a}^{b} f = \int_{[a, b]} f d λ .

This is Lebesgue's criterion for Riemann integrability. For a bounded function, the upper and lower Riemann sums differ by the total oscillation over the partition intervals. The points where the oscillation does not become small are exactly the discontinuities. If that set has outer measure $0$ , cover it by intervals of arbitrarily small total length and make the oscillation small on the complementary closed pieces by compactness; the upper and lower sums become arbitrarily close. Conversely, if the upper and lower sums can be made arbitrarily close, the set where the oscillation is at least $1 / n$ must have outer measure $0$ for each $n$ , and the discontinuity set is their countable union. The equality of the Riemann and Lebesgue integrals then follows by approximating $f$ above and below by step functions and using the Lebesgue integral on intervals.

Lebesgue Integral on an Interval

Suppose that

- \infty \leq a < b \leq \infty

and

f : (a, b) \to ℝ

is Lebesgue measurable. Let

λ

be Lebesgue measure on

ℝ

. Define

\int_{a}^{b} f : = \int_{(a, b)} f d λ

and define

\int_{b}^{a} f : = - \int_{a}^{b} f

L^{1}

-Norm

Suppose that

(X, 𝒮, μ)

is a measure space and

f : X \to [- \infty, \infty]

𝒮

-measurable. Define

{‖ f ‖}_{1} : = \int | f | d μ .

L^{1}

-Space

Suppose that

(X, 𝒮, μ)

is a measure space. The Lebesgue space

L^{1} (μ)

L^{1} (μ) : = {f : X \to ℝ : f is 𝒮 -measurable and {‖ f ‖}_{1} < \infty} .

Approximation in

L^{1}

by Simple Functions

Suppose that

μ

is a measure and

f \in L^{1} (μ)

. For every

ϵ \in ℝ^{+}

, there exists a simple function

g \in L^{1} (μ)

such that

{‖ f - g ‖}_{1} < ϵ .

Apply approximation by simple functions to $f^{+}$ and $f^{-}$ . The monotone convergence theorem implies that the integrals of the approximation errors decrease to $0$ . Choosing sufficiently far along both approximating sequences and subtracting gives a simple function $g$ with ${‖ f - g ‖}_{1} < ϵ$ . Since ${‖ g ‖}_{1} \leq {‖ f ‖}_{1} + {‖ f - g ‖}_{1}$ , we also have $g \in L^{1} (μ)$ .

Step Function

A step function is a function

g : ℝ \to ℝ

of the form

g = a_{1} χ_{I_{1}} + \dots + a_{n} χ_{I_{n}},

where

I_{1}, \dots, I_{n}

are intervals of

ℝ

and

a_{1}, \dots, a_{n} \in ℝ ∖ {0}

Approximation in

L^{1}

by Continuous Functions

Suppose that

f \in L^{1} (ℝ)

. For every

ϵ \in ℝ^{+}

, there exists a continuous function

g : ℝ \to ℝ

such that

{‖ f - g ‖}_{1} < ϵ

and

{x \in ℝ : g (x) \neq 0}

is bounded.

By approximation in $L^{1}$ by simple functions, first approximate $f$ by a finite linear combination of characteristic functions of Lebesgue measurable sets of finite measure. Using the regularity characterization, approximate each measurable set in $L^{1}$ by a finite union of intervals. Finally replace each interval characteristic function by a continuous tent function that is $1$ on most of the interval and $0$ outside a slightly larger interval. The accumulated $L^{1}$ -error can be made less than $ϵ$ , and the resulting continuous function has bounded support.