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

Lebesgue integral (for simple nonnegative measurable functions)

Let $\varphi \in L^{+}(E)$ be a simple, nonnegative measurable function such that $\varphi ={\sum }_{i=1}^n{a}_j{1}_{A_i}$ where $A_i\cap A_j=\varnothing$ for all $i,j$ and ${\bigcup }_{i=1}^n{A}_i=E$. Then the Lebesgue integral of $\varphi$ is ${\int }_E\varphi ={\sum }_{i=1}^n{a}_{i}m(A_i)\in [0,\infty ]$

Now, for general nonnegative measurable functions we have

Lebesgue integral (for general nonnegative measurable functions)

If $f\in L^{+}(E)$, the Lebesgue integral of $f$ is ${\int }_{E}f=\sup \left\{{\int }_E\phi :\phi \in L^{+}(E)\text{ simple, }\phi \le f\right\}$

Theorem

If $E\subset \mathbb{R}$ 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 $\phi$ is a simple function such that $\phi ={\sum }_{i=1}^n{a}_i{\chi }_{A_i}$ and $\phi \le f$, then $m(A_i)\le m(A)=0$. So in the sum, all terms must be $0$. Thus we always have ${\int }_E\phi =0$ and the supremum over all such $\phi$ is also $0$.

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

Note

1. if $\phi \in L^{+}(E)$ is a simple function, then the two definitions of ${\int }_E\phi$ agree
2. If $f,g\in L^{+}(E)$, $c\in [0,\infty ]$, and $f\le g$ on $E$ then

• ${\int }_{E}cf=c{\int }_{E}f$
• ${\int }_{E}f\le {\int }_{E}g$

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

• ${\int }_{F}f={\int }_{E}f{\chi }_F\le {\int }_{E}f$

Theorem

If $f,g\in L^{+}(E)$ and $f\le g$ almost everywhere on $E$ then ${\int }_{E}f\le {\int }_{E}g$

Proof

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

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

\int _{E} f \, &= \int _{F} f + \cancelto{0}{\int _{F^c} f} \\ &= \int _{F} f\, \\ &\leq \int _{F} g \, + \int _{F^c} g \\ &= \int _{E} g \end{align}$$ $$\tag*{$\blacksquare$}$$

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 $f_1\le f_2\le \dots$ 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 $f_1\le f_2\le \dots$, we have that ${\int }_E{f}_1\le {\int }_E{f}_2\le \dots$ (since function relations almost everywhere hold in the integral). Thus ${\left\{{\int }_E{f}_n\right\}}_n$ 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 $f_1\le f_2\le \cdots \le f$

Thus for all $n$, we have $\forall n,{\int }_E{f}_n\le {\int }_{E}f⟹{\lim }_{n\to \infty }{\int }_E{f}_n\le {\int }_{E}f$ Thus to show equality, we have need to show ${\int }_{E}f\le {\lim }_{n\to \infty }{\int }_E{f}_n$.

Now, let $\phi \in L^{+}(E)$ be simple ie $\phi ={\sum }_{i=1}^m{a}_i{\chi }_{A_i}$ with $\phi \le f$. Let $ϵ\in (0,1)$. and define $E_n=\{x\in E|f_n(x)\ge (1-ϵ)\phi (x)\}$ Note that for all $x$, we then have $(1-ϵ)\phi (x)<f(x)$. And since ${\lim }_{n\to \infty }{f}_n(x)=f$, every $x$ must lie in some $E_n$. Thus ${\bigcup }_n{E}_n=E$

Then we have

\int _{E} f_{n } &\geq \int _{E_{n}} f_{n} \\ &\geq \int _{E_{n}} (1-\epsilon) \varphi \\ &= (1-\epsilon)\int _{E} \varphi \\ &= (1-\epsilon) \sum_{i=1}^m a_i m(A_{i} \cap E_{n}) \end{align}$$ And since $E_{1} \cap A_{i} \subset E_{2} \cap A_{i} \subset\dots$ and $\bigcup_{n} (E_{n} \cap A_{i}) = A_{i}$, this implies that $$\begin{align} m(A_{i} \cap E_{n}) \to m(A_{i}) \quad \text{ as }\quad n\to \infty \end{align}$$ Thus $$\begin{align} \lim_{ n \to \infty } \int _{E} f_{n} &\geq \lim_{ n \to \infty } (1-\epsilon) \sum_{i=1}^m a_{i} m(A_{i}\cap E_{n}) \\ &= (1-\epsilon) \sum_{i=1}^m a_{i} m(A_{i}) \\ &= (1-\epsilon) \int _{E} \varphi \end{align}$$ Thus for all $\epsilon \in (0,1)$ we have $$\begin{align} (1-\epsilon) \int _{E} \varphi &\leq \lim_{ n \to \infty } \int _{E} f_{n} \\ \implies_{\epsilon\to {0}} \int _{E} \varphi &\leq \lim_{ n \to \infty } \int _{E} f_{n} \\ &= \int _{E} f \end{align}$$ $$\tag*{$\blacksquare$}$$

(see Monotone Convergence Theorem)

Theorem

Let $f\in L^{+}(E)$ and let $\{{\phi }_n{\}}_n$ be a sequence of simple functions such that $0\le {\phi }_1\le {\phi }_2\le \cdots \le f$ with ${\phi }_n\to f$ pointwise. Then ${\int }_{E}f={\lim }_{n\to \infty }{\int }_E{\phi }_n$

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 ${\phi }_n$ 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 $\{{\phi }_n{\}}_n$ and $\{{\psi }_n{\}}_n$ be two sequences of simple functions such that $0\le {\phi }_1\le {\phi }_2\le \cdots \le f$ with ${\phi }_n\to f$ and $o\le {\psi }_1\le {\psi }_2\le \cdots \le g$ with ${\psi }_n\to g$ pointwise then $0\le {\phi }_1+{\psi }_1\le {\phi }_2+{\psi }_2\le \cdots \le f+g$ Where ${\phi }_n+{\psi }_n\to f+g$ pointwise and each ${\phi }_i+{\psi }_i$ are simple (since they are the sum of simple functions). So by the Monotone Convergence Theorem and linearity for simple functions we have

\int _{E} (f+g) &= \lim_{ n \to \infty } \int _{E} (\varphi_{n} + \psi_{n}) \\ \text{(linearity)}&= \lim_{ n \to \infty } \int _{E} \varphi_{n} + \int _{E} \psi_{n} \\ \text{(MTC)}&=\int _{E} f+\int _{E} g \end{align}$$

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 $N\in \mathbb{N}$ that ${\int }_E{\sum }_{n=1}^N{f}_n={\sum }_{n=1}^N{\int }_E{f}_n$ And since ${\sum }_{n=1}^1{f}_n\le {\sum }_{n=1}^2{f}_n\le \dots$ 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

\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{align}$$ $$\tag*{$\blacksquare$}$$

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 $0\le {\int }_{E}f\le {\int }_{E}0=0$

$(⟹)$ Let $F_n=\left\{x\in E:f(x)>\frac{1}{n}\right\},F=\{x\in E:f(x)>0\}$ Then $F={\bigcup }_n{F}_n$ and $F_1\subset F_2\subset \dots$ and for all $n$ $0\le \frac{1}{n}m(F_n)={\int }_{F_n}\frac{1}{n}\le {\int }_{F_n}f\le {\int }_{E}f=0$ Then $\frac{1}{n}m(F_n)=0⟹m(F_n)=0$ for all $n$. But then by continuity of measure, we have $m(F)=m\left({\bigcup }_{n=1}^{\infty }{F}_n\right)={\lim }_{n\to \infty }m(F_n)=0$ \tag*{$\blacksquare$}

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 $f_1(x)\le f_2(x)\le \dots$ for almost all $x\in E$ and

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 $F=\{x\in E|(\ast )\text{ holds}\}$ Then $m(E\setminus F)=0$ by assumption. Thus $f-{\chi }_{F}f=0$ a.e. and $f_n-{\chi }_F{f}_n=0$ a.e. also. Then the Monotone Convergence Theorem says ${\int }_{E}f={\int }_{E}f{\chi }_F={\int }_{F}f={\lim }_{n\to \infty }{\int }_F{f}_n$ 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 $m(E\setminus F)=0$ 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 ${\int }_E{\lim }_{n\to \infty }\inf f_n(x)dx\le {\lim }_{n\to \infty }\inf \int f_n(x)dx$

Recall ${\lim }_{n\to \infty }\inf a_n={\sup }_{n\ge 1}[{\inf }_{k\ge n}{a}_k]$ And the ${\lim }_{n\to \infty }\inf$ function is defined pointwise

Proof

We have

\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{align}$$ And since $\inf_{k \geq 1} f_{k}(x) \leq \inf_{k \geq 2} f_{k} \leq\dots$ by the [[Monotone Convergence Theorem|MCT]] we have $$\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{align} \inf_{k \geq n} f_{k}(x) &\leq f_{j} (x) \\ \implies \int _{E} \inf_{k \geq n} f_{k} &\leq \int _{E} f_{j} \\ \implies \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 } \left[\inf_{j \geq n} \int _{E} f_{j}\right] \\ &= \lim_{ n \to \infty } \inf \int _{E}f_{n} \end{align}$$ So we have "swapped the integral and inf" and we can plug this into the MCT to get the desired result. $$\tag*{$\blacksquare$}$$

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 $n\in \mathbb{N}$ we have $n{\chi }_F\le f{\chi }_F$ So integrating both sides gives us $nm(F)\le {\int }_{E}f{\chi }_F\le {\int }_{E}f\le \infty$ Thus for all $n$, we have $m(F)\le \frac{1}{n}{\int }_{E}f$. 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.

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