Functional Analysis Lecture 11

[[lecture-data]]

← Lecture 10 | Lecture 12 →
Lecture Notes: Rodriguez, page 53

2. Lebesgue Measure and Integration

[[Lebesgue Integral]] of nonnegative functions

Recall from last time

L^{+}

functions

If $E \subset R$ is measurable, the class of nonnegative measurable functions on $E$ is given as
$L^{+} (E) = {f : E \to [0, \infty] | f is measurable}$

)

Let $ϕ \in L^{+} (E)$ be a simple, nonnegative measurable function such that $ϕ = \sum_{i = 1}^{n} a_{j} 1_{A_{i}}$ where $A_{i} \cap A_{j} = \emptyset$ for all $i, j$ and $⋃_{i = 1}^{n} A_{i} = E$ . Then the Lebesgue integral of $ϕ$ is
$\int_{E} ϕ = \sum_{i = 1}^{n} a_{i} m (A_{i}) \in [0, \infty]$

Now, for general [[nonnegative measurable functions]] we have

)

If $f \in L^{+} (E)$ , the Lebesgue integral of $f$ is

\int_{E} f = sup {\int_{E} φ : φ \in L^{+} (E) simple, φ \leq f}

Theorem

If $E \subset R$ is a set with $m (E) = 0$ , then for all $f \in L^{+} (E)$ we have $\int_{E} f = 0$

ie it is only interesting to take integrals over functions with positive measure

Proof

From the definition, we have $f \in L^{+} (E)$ . If $φ$ is a [[simple function]] such that $φ = \sum_{i = 1}^{n} a_{i} χ_{A_{i}}$ and $φ \leq f$ , then $m (A_{i}) \leq m (A) = 0$ . So in the sum, all terms must be $0$ . Thus we always have $\int_{E} φ = 0$ and the supremum over all such $φ$ is also $0$ .

see [[it is only interesting to take integrals of functions with positive measure]]

Note

if $φ \in L^{+} (E)$ is a [[simple function]], then the two definitions of $\int_{E} φ$ agree
If $f, g \in L^{+} (E)$ , $c \in [0, \infty]$ , and $f \leq g$ on $E$ then

$\int_{E} c f = c \int_{E} f$
$\int_{E} f \leq \int_{E} g$

If $f \in L^{+} (E)$ and $F \subset E$ is measurable, then

$\int_{F} f = \int_{E} f χ_{F} \leq \int_{E} f$

Theorem

If $f, g \in L^{+} (E)$ and $f \leq g$ [[almost everywhere]] on $E$ then

\int_{E} f \leq \int_{E} g

Proof

Let $F = {x \in E : f (x) \leq g (x)}$ .

This is measurable since $g - f$ is measurable (sums of measurable functions are measurable)
And $m (F^{c}) = 0$ . Then

\begin{aligned} \int_{E} f & = \int_{F} f + {\int_{F^{c}} f}^{0} \\ = \int_{F} f \\ \leq \int_{F} g + \int_{F^{c}} g \\ = \int_{E} g \end{aligned}

\begin{matrix} ◼ \end{matrix}

See [[function relations almost everywhere hold in the integral]]

Monotone Convergence Theorem

If ${f_{n}}_{n}$ is a sequence in $L^{+} (E)$ (nonnegative and measurable) such that $f_{1} \leq f_{2} \leq \dots$ and $f_{n} \to f$ pointwise on $E$ , then

lim_{n \to \infty} \int_{E} f_{n} = \int_{E} f

Note

Pointwise convergence is much weather than uniform convergence required for Reimann integration

Proof

Since $f_{1} \leq f_{2} \leq \dots$ , we have that $\int_{E} f_{1} \leq \int_{E} f_{2} \leq \dots$ (since [[function relations almost everywhere hold in the integral]]). Thus ${\int_{E} f_{n}}_{n}$ is an increasing sequence on nonnegative numbers. Thus $lim_{n \to \infty} \int_{E} f_{n} \in [0, \infty]$ .

And since $lim_{n \to \infty} f_{n} (x) = f (x)$ for all $x \in E$ , we know $f_{1} \leq f_{2} \leq \dots \leq f$

Thus for all $n$ , we have

\forall n, \int_{E} f_{n} \leq \int_{E} f ⟹ lim_{n \to \infty} \int_{E} f_{n} \leq \int_{E} f

Thus to show equality, we have need to show $\int_{E} f \leq lim_{n \to \infty} \int_{E} f_{n}$ .

Now, let $φ \in L^{+} (E)$ be simple ie $φ = \sum_{i = 1}^{m} a_{i} χ_{A_{i}}$ with $φ \leq f$ . Let $ϵ \in (0, 1)$ . and define

E_{n} = {x \in E | f_{n} (x) \geq (1 - ϵ) φ (x)}

Note that for all $x$ , we then have $(1 - ϵ) φ (x) < f (x)$ . And since $lim_{n \to \infty} f_{n} (x) = f$ , every $x$ must lie in some $E_{n}$ . Thus $⋃_{n} E_{n} = E$

Then we have

\begin{aligned} \int_{E} f_{n} & \geq \int_{E_{n}} f_{n} \\ \geq \int_{E_{n}} (1 - ϵ) φ \\ = (1 - ϵ) \int_{E} φ \\ = (1 - ϵ) \sum_{i = 1}^{m} a_{i} m (A_{i} \cap E_{n}) \end{aligned}

And since $E_{1} \cap A_{i} \subset E_{2} \cap A_{i} \subset \dots$ and $⋃_{n} (E_{n} \cap A_{i}) = A_{i}$ , this implies that

\begin{array}{r} m (A_{i} \cap E_{n}) \to m (A_{i}) as n \to \infty \end{array}

Thus

\begin{aligned} lim_{n \to \infty} \int_{E} f_{n} & \geq lim_{n \to \infty} (1 - ϵ) \sum_{i = 1}^{m} a_{i} m (A_{i} \cap E_{n}) \\ = (1 - ϵ) \sum_{i = 1}^{m} a_{i} m (A_{i}) \\ = (1 - ϵ) \int_{E} φ \end{aligned}

Thus for all $ϵ \in (0, 1)$ we have

\begin{aligned} (1 - ϵ) \int_{E} φ & \leq lim_{n \to \infty} \int_{E} f_{n} \\ ⟹_{ϵ \to 0} \int_{E} φ & \leq lim_{n \to \infty} \int_{E} f_{n} \\ = \int_{E} f \end{aligned}

\begin{matrix} ◼ \end{matrix}

(see [[Monotone Convergence Theorem]])

Theorem

Let $f \in L^{+} (E)$ and let ${φ_{n}}_{n}$ be a sequence of simple functions such that

0 \leq φ_{1} \leq φ_{2} \leq \dots \leq f

with $φ_{n} \to f$ pointwise. Then

\int_{E} f = lim_{n \to \infty} \int_{E} φ_{n}

ie, we can take any pointwise increasing sequence of simple functions and compute the limit (instead of needed to compute the supremum)

Proof

We can just plug the $φ_{n}$ in to the [[Monotone Convergence Theorem]]

Theorem

If $f, g \in L^{+} (E)$ then

\int_{E} (f + g) = \int_{E} f + \int_{E} g

Proof

Let ${φ_{n}}_{n}$ and ${ψ_{n}}_{n}$ be two sequences of simple functions such that
$0 \leq φ_{1} \leq φ_{2} \leq \dots \leq f$ with $φ_{n} \to f$ and
$o \leq ψ_{1} \leq ψ_{2} \leq \dots \leq g$ with $ψ_{n} \to g$ pointwise then

0 \leq φ_{1} + ψ_{1} \leq φ_{2} + ψ_{2} \leq \dots \leq f + g

Where $φ_{n} + ψ_{n} \to f + g$ pointwise and each $φ_{i} + ψ_{i}$ are simple (since they are the sum of simple functions). So by the [[Monotone Convergence Theorem]] and linearity for simple functions we have

\begin{aligned} \int_{E} (f + g) & = lim_{n \to \infty} \int_{E} (φ_{n} + ψ_{n}) \\ (linearity) & = lim_{n \to \infty} \int_{E} φ_{n} + \int_{E} ψ_{n} \\ (MTC) & = \int_{E} f + \int_{E} g \end{aligned}

See [[lebesgue integral of sum is sum of the integral]]

Theorem

Let ${f_{n}}_{n}$ be a sequence in $L^{+} (E)$ . Then

\int_{E} \sum_{n} f_{n} = \sum_{n} \int_{E} f_{n}

Proof

via induction. Since the [[lebesgue integral of sum is sum of the integral]], we know for each $N \in N$ that

\int_{E} \sum_{n = 1}^{N} f_{n} = \sum_{n = 1}^{N} \int_{E} f_{n}

And since

\sum_{n = 1}^{1} f_{n} \leq \sum_{n = 1}^{2} f_{n} \leq \dots

and $\sum_{n = 1}^{N} f_{n} \to \sum_{n = 1}^{\infty} f_{n}$ as $N \to \infty$ , by the [[Monotone Convergence Theorem]] we have

\begin{aligned} \int_{E} \sum_{n = 1}^{\infty} & = lim_{N \to \infty} \int_{E} \sum_{n = 1}^{N} f_{n} \\ = lim_{N \to \infty} \sum_{n = 1}^{N} \int_{E} f_{n} \\ = \sum_{n = 1}^{\infty} \int_{E} f_{n} \end{aligned}

\begin{matrix} ◼ \end{matrix}

see [[integral of sum of a sequence is sum of the integrals]]

Note

this does not hold for Riemann integration. Enumerate the rationals and let $f_{n}$ be the function that is 1 for the first $n$ rational numbers and 0 everywhere else

Theorem

Let $f \in L^{+} (E)$ . Then $\int_{E} f = 0 ⟺ f = 0$ [[almost everywhere]] on $E$

Proof

(⟸)

If $f = 0$ [[almost everywhere]]

0 \leq \int_{E} f \leq \int_{E} 0 = 0

(⟹)

Let

F_{n} = {x \in E : f (x) > \frac{1}{n}}, F = {x \in E : f (x) > 0}

Then $F = ⋃_{n} F_{n}$ and $F_{1} \subset F_{2} \subset \dots$ and for all $n$

0 \leq \frac{1}{n} m (F_{n}) = \int_{F_{n}} \frac{1}{n} \leq \int_{F_{n}} f \leq \int_{E} f = 0

Then $\frac{1}{n} m (F_{n}) = 0 ⟹ m (F_{n}) = 0$ for all $n$ . But then by continuity of measure, we have

m (F) = m (⋃_{n = 1}^{\infty} F_{n}) = lim_{n \to \infty} m (F_{n}) = 0

\begin{matrix} ◼ \end{matrix}

see [[integral is 0 if and only if the function is 0 almost everywhere]]

Theorem

If ${f_{n}}_{n}$ is a sequence in $L^{+} (E)$ such that $f_{1} (x) \leq f_{2} (x) \leq \dots$ for almost all $x \in E$ and

(*) {\begin{cases} f_{1} (x) \leq f_{2} (x) \leq f_{3} (x) \dots \\ lim_{n \to \infty} f_{n} (x) = f (x) \end{cases}

Then

\int_{E} f = lim_{n \to \infty} \int_{E} f_{n}

Proof

Let

F = {x \in E | (*) holds}

Then $m (E ∖ F) = 0$ by assumption. Thus $f - χ_{F} f = 0$ a.e. and $f_{n} - χ_{F} f_{n} = 0$ a.e. also. Then the [[Monotone Convergence Theorem]] says

\int_{E} f = \int_{E} f χ_{F} = \int_{F} f = lim_{n \to \infty} \int_{F} f_{n}

Where the first equality holds since [[function relations almost everywhere hold in the integral]] and the last holds by the [[Monotone Convergence Theorem]]. This then becomes

lim_{n \to \infty} \int_{F} f_{n} = lim_{n \to \infty} \int_{E} f_{n}

Because $m (E ∖ F) = 0$ and so any integral over the region is 0.

ie, sets of measure zero do not affect the [[Lebesgue Integral]].

see [[sets of measure zero do not affect the integral]]

Theorem (Fatou's Lemma)

Let ${f_{n}}_{n}$ be a sequence in $L^{+} (E)$ . Then

\int_{E} lim_{n \to \infty} inf f_{n} (x) d x \leq lim_{n \to \infty} inf \int f_{n} (x) d x

Recall

$lim_{n \to \infty} inf a_{n} = sup_{n \geq 1} [inf_{k \geq n} a_{k}]$

And the $lim_{n \to \infty} inf$ function is defined pointwise

Proof

We have

\begin{aligned} lim_{n \to \infty} inf f_{n} (x) & = sup_{n \geq 1} [inf_{k \geq n} f_{k} (x)] \\ = lim_{n \to \infty} [inf_{k \geq n} f_{k} (x)] \end{aligned}

And since $inf_{k \geq 1} f_{k} (x) \leq inf_{k \geq 2} f_{k} \leq \dots$ by the MCT we have

\int_{E} lim_{n \to \infty} inf f_{n} = lim_{n \to \infty} \int_{E} inf_{k \geq n} f_{k}

For all $j \geq n$ and all $x \in E$ , we have

\begin{aligned} inf_{k \geq n} f_{k} (x) & \leq f_{j} (x) \\ ⟹ \int_{E} inf_{k \geq n} f_{k} & \leq \int_{E} f_{j} \\ ⟹ \int_{E} lim_{n \to \infty} inf f_{n} & = lim_{n \to \infty} \int_{E} (inf_{k \geq n} f_{k}) \\ \leq lim_{n \to \infty} [inf_{j \geq n} \int_{E} f_{j}] \\ = lim_{n \to \infty} inf \int_{E} f_{n} \end{aligned}

So we have "swapped the integral and inf" and we can plug this into the MCT to get the desired result.

\begin{matrix} ◼ \end{matrix}

see [[Fatou's Lemma]]

Theorem

Let $f \in L^{+} (E)$ and let $\int_{E} f < \infty$ . Then the set

F = {x \in E : f (x) = \infty}

must have measure 0.

Proof

We know for all $n \in N$ we have

n χ_{F} \leq f χ_{F}

So integrating both sides gives us

n m (F) \leq \int_{E} f χ_{F} \leq \int_{E} f \leq \infty

Thus for all $n$ , we have $m (F) \leq \frac{1}{n} \int_{E} f$ . Taking $n \to \infty$ , we have $m (F) = 0$ .

see [[functions with finite integrals map a measure zero set to infinity]]

Next things:

define set of all [[Lebesgue integrable]] functions
Show that they are a normed space, build up a [[Banach space]] and $L^{p}$ spaces.

const { dateTime } = await cJS()

return function View() {
	const file = dc.useCurrentFile();
	return <p class="dv-modified">Created {dateTime.getCreated(file)}     ֍     Last Modified {dateTime.getLastMod(file)}</p>
}