Functional Analysis Lecture 8

[[lecture-data]]

← Lecture 7 | Lecture 9 → Lecture Notes: Rodriguez, page 36

2. Lebesgue measure and integration

Recall from last time

Algebra

A nonempty collection of sets $\mathcal{A}\subset \mathcal{P}(\mathbb{R})$ is an algebra if

1. (closure under complements) $E\in \mathcal{A}⟹E^c\in \mathcal{A}$
2. (closure under finite unions)$E_1,\dots ,E_n\in \mathcal{A}⟹{\bigcup }_{k=1}^n{E}_k\in \mathcal{A}$

[!def] $\sigma -$algebra

We say an algebra is a $\sigma -$algebra (sigma-algebra) if also 3. (closure under countable unions) $\{E_n{\}}_{n=1}^{\infty }\in \mathcal{A}⟹{\bigcup }_n{E}_n\in \mathcal{A}$

Lebesgue Measurable

A set $E\subset \mathbb{R}$ is Lebesgue measurable if for all $A\subset \mathbb{R}$ we have $m^(A) = m^(A \cap E) + m^*(A \cap E^c)$$

Today we will

• show that the borel sigma algebra is a subset of all measurable sets $\mathcal{M}$
• define Lebesgue measure finally

Lemma

Let $\mathcal{A}$ be an algebra and let $\{E_n\}$ be a countable collection of elements in $\mathcal{A}$. Then there exists $\{F_n\}$ a disjoint countable collection of elements in $\mathcal{A}$ such that ${\bigcup }_n{E}_n={\bigcup }_n{F}_n$

ie we can get closure under countable unions when we have closure under countable disjoint unions

Proof

Let $G_n={\bigcup }_{k=1}^n{E}_k$ so that $G_1\subset G_2\subset G_3\subset \dots$ and ${\bigcup }_n{G}_n={\bigcup }_n{E}_n$. Now, let $F_1=G_1$ and for all $k$ define $F_{k+1}=G_{k+1}\setminus G_k$ Then we also have that the disjoint ${\bigcup }_n{F}_n={\bigcup }_n{G}_n={\bigcup }_n{E}_n$.
\tag*{$\blacksquare$}

see algebras have closure under finite disjoint countable unions

Theorem

Let $A\subset \mathbb{R}$ and let $E_1,\dots ,E_n$ be disjoint and measurable. Then $m^{\ast }\left(A\cap \left[{\bigcup }_{k=1}^n{E}_k\right]\right)={\sum }_{k=1}^n{m}^{\ast }(A\cap E_k)$

NOTE

if $E_1=E$ and $E_2=E^c$, this is the definition of measurability.

Proof

Via induction. Trivially true for $n=1$. Suppose that the equality is true for $n=m$. Now, suppose we have pairwise disjoint measurable sets $E_1,\dots ,E_{m+1}$ and $A\subset \mathbb{R}$. Since $E_k\cap E_{m+1}=\varnothing$ for all $k$, we have (since $E_{m+1}$ is measurable ) that

A \cap \left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E_{m+1} &= A \cap E_{m+1} \\ \text{and}\quad A \cap\left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E^c_{m+1} &= A \cap \left[ \bigcup_{k=1}^m E_{k} \right] \\ \implies m^*\left( A \cap \left[ \bigcup_{k=1}^{m+1}E_{k} \right] \right) &= m^*\left( A \cap \left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E_{m+1} \right) + m^*\left( A \cap\left[ \bigcup_{k=1}^{m+1} E_{k} \right] \cap E^c_{m+1} \right) \\ &= m^*(A \cap E_{m+1} ) + m^*\left(A \cap \left[ \bigcup_{k=1}^m E_{k} \right] \right) \\ &= m^*(A \cap E_{m+1}) + \sum_{k=1}^m m^*(A \cap E_{k}) \end{align}$$ Where the last inequality is due to the induction hypothesis. Thus the result follows via induction. $$\tag*{$\blacksquare$}$$

see measure of finite disjoint measurable sets is the sum of the measures

Theorem

The collection of measurable sets $\mathcal{M}$ is a sigma-algebra

Proof

We already know that $\mathcal{M}$ is an algebra, and we know since algebras have closure under finite disjoint countable unions that we just need to check countable disjoint unions are measurable.

So let $\{E_n\}$ be a countable collection of disjoint measurable sets with $E={\bigcup }_n{E}_n$. Then since the other half is always satisfied via monotonicity, we just need to show $m^{\ast }(A\cap E^c)+m^{\ast }(A\cap E)\le m^{\ast }(A)$

Let $N\in \mathbb{N}$. Since $\mathcal{M}$ is an algebra, the union ${\bigcup }_{n=1}^N{E}_n\in \mathcal{M}$ so we have $m^{\ast }(A)=m^{\ast }\left(A\cap \left[{\bigcup }_{n=1}^N{E}_n\right]\right)+m^{\ast }\left(A\cap {\left[{\bigcup }_{n=1}^N{E}_n\right]}^c\right)$ And since ${\bigcup }_{n=1}^N{E}_n\subset E$, we have $E^c\subset {\left[{\bigcup }_{n=1}^N{E}_n\right]}^c$. Thus we have

m^*(A) &= m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n} \right] \right)+m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n}\right]^c \right) \\ &\geq m^*\left( A \cap \left[ \bigcup_{n=1}^N E_{n} \right] \right) + m^*(A \cap E^c) \end{align}$$ And since the [[measure of finite disjoint measurable sets is the sum of the measures]], we get $$\begin{align} m^*(A) &\geq \sum_{n=1}^N m^*(A \cap E_{n}) + m^*(A \cap E^c) \\ (N\to \infty )&\geq \sum_{n=1}^\infty m^*(A \cap E_{n}) + m^*(A \cap E^c) \\ (*) &\geq m^*\left( \bigcup_{n} A \cap E_{n} \right) + m^*(A \cap E^c) \\ &= m^*(A \cap E) + m^*(A \cap E^c) \end{align}$$ Where $(*)$ is from [[outer measure has countable subadditivity|countable subadditivity]]. $$\tag*{$\blacksquare$}$$

See measurable sets form a sigma algebra

NOTE

We want measurable sets to form a sigma algebra based on the desirable properties for measure:

Properties we want for this idea

1. $m(E)$ is defined for all $E\subset \mathbb{R}$

2. if $I$ is an interval then $m(E)=\ell (I)$ the “length” of $I$

3. If $\{E_n\}$ is a (countable) collection of disjoint subsets of $E$ such that $E={\bigcup }_n{E}_n$, then we want $m\left({\bigcup }_n{E}_n\right)=m(E)$

4. Translation invariance. ie, if $E\subset \mathbb{R}$ and $x\in \mathbb{R}$, then $m(x+E)=m(\{x+y|y\in E\})=m(E)$

In particular, in order to get (3), we need to be able to define measure on any countable disjoint union.

Since we know the measurable sets form a sigma algebra, we now can show that it contains $\mathcal{B}$ the borel sigma algebra. To do this, we just need to show that it contains all open sets.

Proposition

For all $a\in \mathbb{R}$, the interval $(a,\infty )$ is measurable.

Proof

Suppose $A\subset \mathbb{R}$. Let $A_1=A\cap (a,\infty )$ and $A_2=A\cap (-\infty ,a]$. Now we want to show that $m^{\ast }(A_1)+m^{\ast }(A_2)\le m^{\ast }(A)$.

If $m^{\ast }(A)$ is infinite we are done. So suppose that $m^{\ast }(A)<\infty$. Now, let $\{I_n\}$ be a collection of intervals such that ${\sum }_n\ell (I_n)\le m^{\ast }(A)+\epsilon$ And define $J_n=I_n\cap (a,\infty )K_n=I_n\cap (-\infty ,a]$ Then for each $n$, each of $J_n,K_n$ are either an interval are empty and

• $A_1\subset {\bigcup }_n{J}_n$
• $A_2\subset {\bigcup }_n{K}_n$
• $\ell (I_n)=\ell (J_n)+\ell (K_n)$ Thus we have

m^*(A_{1}) + m^*(A_{2}) &\leq \sum_{n} m^*(J_{n}) + m^*(K_{n}) \\ &= \sum_{n} \ell(J_{n}) + \ell(K_{n}) \\ &= \sum_{n} \ell(I_{n}) \\ &\leq m^*(A) + \varepsilon \end{align}$$ Then take $\varepsilon \to 0$ and we have the desired result. $$\tag*{$\blacksquare$}$$

see open intervals with upper bound infinity are measurable

Theorem

Every open subset of $\mathbb{R}$ is measurable (ie $\mathcal{B}\subset \mathcal{M}$ - the borel sigma algebra is contained in the collection of all measurable sets)

Proof

Since intervals of the form $(a,\infty )$ are measurable for all $a\in \mathbb{R}$ (open intervals with upper bound infinity are measurable), so is $(-\infty ,b)={\bigcup }_{n=1}^{\infty }\left(-\infty ,b-\frac{1}{n}\right]={\bigcup }_{n=1}^{\infty }{\left(b-\frac{1}{n},\infty \right)}^c$ This is because the measurable sets form a sigma algebra and they are therefore closed under complements, countable unions, and finite intersections. Thus any finite open interval is also measurable since $(a,b)=(-\infty ,b)\cap (a,\infty )$ Finally, every open subset of $\mathbb{R}$ is a countable union of open intervals. Thus all open intervals are measurable. \tag*{$\blacksquare$}

see all borel sets are measurable

Lebesgue Measure

The Lebesgue measure of a measurable set $E\subset \mathcal{M}$ is given by $m(E)=m^{\ast }(E)$

see Lebesgue measure

This means we have restricted our notion of measure to only the well-behaved sets (recall the spoiler for our desirable properties). We now need to check countable additivity

Theorem

Suppose that $\{E_n\}$ is a countable collection of disjoint, measurable sets. Then $m\left({\bigcup }_n{E}_n\right)={\sum }_{n}m(E_n)$

countable subadditivity of measure, but now we are showing equality for our nice sets.

We already showed

Proof

We know that the set ${\cup }_n{E}_n$ is measurable since measurable sets form a sigma algebra and are therefore closed under countable unions. So we already have (from countable subadditivity of measure) that $m\left({\bigcup }_n{E}_n\right)=m^{\ast }\left({\bigcup }_n{E}_n\right)\le {\sum }_n{m}^{\ast }(E_n)={\sum }_{n}m(E_n)$ So we just need to show the other way. For any $N\in \mathbb{N}$, since measure of finite disjoint measurable sets is the sum of the measures, we have

m\left( \bigcup_{n=1}^N E_{n} \right) &= m^*\left( \mathbb{R} \cap \bigcup_{n=1}^N E_{n} \right) \\ &= \sum_{n=1}^N m^*(\mathbb{R} \cap E_{n}) \\ &= \sum_{n=1}^N m^*(E_{n}) \\ &= \sum_{n=1}^N m(E_{n}) \\ (*)&= m\left( \bigcup_{n=1}^N E_{n}\right) \\ &\leq m\left( \bigcup_{n=1}^\infty E_{n} \right) \end{align}$$ Where $(*)$ is because of the disjointness of the $E_{n}$. Now we have a bound over all $N$, and taking $N \to \infty$ we get the desired result. $$\tag*{$\blacksquare$}$$

see measure satisfies countable additivity

The final condition that we need to verify from our desirable properties for measure are translation invariance. That is, if $E\in \mathcal{M}$ and $x\in \mathbb{R}$ then $m(E+x)=m(E)$ (where $E+x=\{y+x:y\in E\}$ ) (HW4, p3a)

Theorem

Suppose $\{E_k\}$ is a countable collection of measurable sets such that $E_1\subset E_2\subset \dots$ Then $m\left({\bigcup }_{k=1}^{\infty }{E}_k\right)={\lim }_{n\to \infty }m\left({\bigcup }_{k=1}^n\right)={\lim }_{n\to \infty }m(E_n)$

Proof

The second equality is because $E_n={\bigcup }_{k=1}^n{E}_k$. So it is enough to just show $m\left({\bigcup }_k{E}_k\right)={\lim }_{n\to \infty }m(E_n)$.

We can do this by showing the countable union as the countable disjoint union (recall that algebras have closure under finite disjoint countable unions).

So define $F_1=E_1$ and $F_k=E_k\setminus E_{k-1}$. Each $F_k$ is measurable since $F_k=E_k\cap E_{k-1}^c$ and the collection $\{F_k\}$ is disjoint. Then for all $n\in \mathbb{N}$, we have

\bigcup_{k=1}^n F_{k} = E_{n}\, \quad&\quad \bigcup_{k=1}^\infty F_{k} = \bigcup_{k=1}^\infty E_{k} \\ \implies m\left( \bigcup_{k=1}^\infty E_{k} \right) &= \sum_{k=1}^\infty m(F_{k}) \\ &= \lim_{ n \to \infty } \sum_{k=1}^n m(F_{k}) \\ &= \lim_{ n \to \infty } m\left( \bigcup_{k=1}^n F_{k} \right) \\ &= \lim_{ n \to \infty } m(E_{n}) \end{align}$$ Since [[measure of finite disjoint measurable sets is the sum of the measures]]. Thus we have shown the desired equality. $$\tag*{$\blacksquare$}$$

See measure of union of nested sets converges to measure of limiting set

Next time: Lebesgue measure to define measurable functions - the analog of continuous functions for Riemann integration.

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>
}