Functional Analysis Lecture 7

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← Lecture 6 | Lecture 8 → Lecture Notes: Rodriguez, page 31

2. Lebesgue Measure and Integration

Recall we defined

Outer Measure

For $A\subset \mathbb{R}$, we define the outer measure of $A$ m^*(A) = \inf\left\{ \sum_{n} \ell(I_{n}) : \{ I_{n} \} \text{ countable and open s.t. } A \subset \bigcup_{n}I_{n} \right\}$$ We can see that m^*(A) \geq 0$forall$A$.

Theorem

If $I$ is an interval, then $m^{\ast }(I)=\ell (I)$

Proof

Suppose $I=[a,b]$. Then $I\subset (a-ϵ,b+ϵ)$ for all $ϵ>0$. But then we have $m^{\ast }(I)\le \ell (a-ϵ,b+ϵ)=b-a+2ϵ$ Thus we have $m^{\ast }(I)\le b-a$.

Now we show $b-a\le m^{\ast }(I)$. Let $\{I_n{\}}_n$ be a collection of open intervals such that $[a,b]\subset {\bigcup }_n{I}_n$.

Now, since $[a,b]$ is compact (by the Heine-Borel theorem), there exists a finite subcover $\{J_k{\}}_{k=0}^m\subset \{I_n{\}}_n$ such that $[a,b]\subset {\bigcup }_{k=0}^m{J}_k$

Since $a\in {\bigcup }_{k=0}^m{J}_k$, there exists $k_1$ such that $a\in J_{k_1}$. Rearranging intervals, I let $k_1=1$ and $a\in J_1=(a_1,b_1)$. If $b_1<b$, then $b_1\in [a,b]$. So ther eis some $k_2$ such that $b_1\in J_{k_2}$. By rearranging again, I assume $k_2=2$. So $b_1\in J_2=(a_2,b_2)$.

Continuing in the same manner, we conclude that there exists a $K,1\le K\le m$ such that $b<b_K$. And for all $k\in \{1,2,\dots ,K-1\}$ we have $b_k\le b$ and $a_{k+1}\le b_k<b_{k+1}$.

But then we have

\sum_{n} \ell(I_{n}) & \geq \sum_{k=1}^m \ell(J_{k}) \\ & \geq \sum_{k=1}^K \ell(J_{k}) \\ &= (b_{K}-a_{K}) + (b_{K-1} - a_{K-1}) + \dots + (b_{1}-a_{1}) \\ &= b_{K} + (b_{K-1} - a_{K}) + (b_{K-2}-a_{K-1}) + \dots + (b_{1}-a_{2}) - a_{1} \\ & \geq b_{K} - a_{1} \\ &\geq b-a \\ &= \ell(I) \end{align}$$ Thus we have $m^*(I) \geq \ell(I) = b-a$ Thus we conclude $m^*(I) = \ell(I)$ $$\tag*{$\blacksquare$}$$

See the outer measure of an interval is its length

NOTE

If $I$ is any finite interval $[a,b],[a,b),(a,b],$ or $(a,b)$, we note that for all $ϵ>0$ we have

[a+\epsilon, b-\epsilon] &\subset I &\subset [a-\epsilon, b +\epsilon] \\ \implies m^*([a+\epsilon, b-\epsilon]) &\leq m^*(I) &\leq m^*([a-\epsilon, b +\epsilon]) \\ \implies (b-a) - 2\epsilon &\leq m^*(I) & \leq (b-a) + 2\epsilon \end{align}$$ Taking $\epsilon\to0$, we get the desired result.

Exercise

An infinite interval cannot be covered by a collection of intervals of finite length

Theorem

for every $A\subset \mathbb{R}$ and $ϵ>0$, there exist an open set $O$ such that $A\subset O$ and $m^{\ast }(A)\le m^{\ast }(O)\le m^{\ast }(A)+ϵ$

Proof

The result is obvious if $m^{\ast }(A)=\infty$ (just take $O=\mathbb{R}$), so we assume $m^{\ast }(A)<\infty$.

Let $\{I_n\}$ be a collection of open intervals that cover $A$ and have total length at most $m^{\ast }(A)+ϵ$. Then $O={\bigcup }_n{I}_n$ is open and obviously $A\subset O$. Thus by subadditivity ( theorem from lecture 6 )

m^*(O) &= m^*\left( \bigcup_{n}I_{n} \right) \\ &\leq \sum_{n} m^*(I_{n}) \\ &\leq \sum_{n} \ell(I_{n}) \\ & \leq m^*(A) + \epsilon \end{align}$$

see we can always find an open set with outer measure slightly more

Measurable Sets

Lebesgue measurable

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

NOTE

Since for all $A,E$ we have $A=(A\cap E)\cup (A\cap E^c)$, we have by subadditivity that $m^{\ast }(A)\le m^{\ast }(A\cap E)+m^{\ast }(A\cap E^c)$.

Thus $E$ is measurable if $m^{\ast }(A\cap E)+m^{\ast }(A\cap E^c)\le m^{\ast }(A)$

( we only need to show one side of the inequality )

lebesgue measurable

Theorem

The empty set $\varnothing$ and the set of real numbers $\mathbb{R}$ are measurable. And a set $E$ is measurable if and only if $E^c$ is measurable.

Proof

These are readily verifiable from the definition of measurability - which is symmetric in $E$ and $E^c$.

Proposition

If $m^{\ast }(E)=0$ then $E$ is measurable.

Proof

Let $A\subset \mathbb{R}$. Then $A\cap E\subset E$ means $m^{\ast }(A\cap E)\le m^{\ast }(E)=0⟹m^{\ast }(A\cap E)=0$ But then we see $m^{\ast }(A\cap E)+m^{\ast }(A\cap E^c)=m^{\ast }(A\cap E^c)\le m^{\ast }(A)$ because $A\cap E^c\subset A$.

see outer measure zero sets are measruable

NOTE

Lots of “uninteresting” sets are measurable - it is hard to find interesting sets that are measurable.

• every open set is measurable
• so every closed set is also measurable (by taking complements)
• Most things we can write down are measurable (since taking unions and intersections of basic sets also give us measurable sets)

Proposition

If $E_1,E_2$ are measurable, then $E_1\cup E_2$ are measurable

Proof

Let $A\subset \mathbb{R}$. Since $E_2$ is measurable, we know $m^{\ast }(A\cap E_1^c)=m^{\ast }(A\cap E_1^c\cap E_2)+m^{\ast }(A\cap \underset{(E_1\cup E_2{)}^c}{\underbrace{E_1^c\cap E_2^c}})$ Then we also know

A \cap(E_{1} \cup E_{2}) &= (A \cap E_{1}) \cup (A \cap E_{2}) \\ &= (A \cap E_{1}) \cup (A \cap E_{2} \cap E_{1}^c) \\ \end{align}

Taking outer measure we get

\implies m^*(A\cap(E_{1} \cup E_{2})) & \leq m^*(A \cap E_{1}) + m^*(A \cap E_{2} \cap E_{1}^c) \\ E_{1} \text{measurable } \implies&= m^*(A) - m^*(A \cap E_{1}^c) + m^*(A \cap E_{2} \cap E_{1}^c) \\ &= m^*(A) - m^*(A \cap (E_{1} \cup E_{2})^c) \end{align}$$ And this gives us the desired result.

see unions of measurable sets are measurable

Theorem

If $E_1,\dots ,E_n$ are measurable, then ${\bigcup }_{k=1}^n{E}_k$ is measurable

Proof

by induction. $n=1$ is trivial. Suppose the statement holds for $n=k$. Then ${\bigcup }_{j=1}^{k+1}{E}_j=\left({\bigcup }_{j=1}^k{E}_j\right)\cup E_{k+1}$ And then since unions of measurable sets are measurable we get the desired result.

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)$E_1,\dots ,E_n\in \mathcal{A}⟹{\bigcup }_{k=1}^n{E}_k\in \mathcal{A}$

see algebra, sigma-algebra

$\sigma -$algebra

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

NOTE

By DeMorgan’s Laws, if $E_1,\dots ,E_n\in \mathcal{A}$, then we also have ${\bigcap }_{k=1}^n{E}_k={\left({\bigcup }_{k=1}^n{E}_n\right)}^c\in \mathcal{A}$ Thus, if $E\in \mathcal{A}$ then $\varnothing =E\cap E^c\in \mathcal{A}⟹\mathbb{R}={\varnothing }^c\in \mathcal{A}$ Similarly, if $\mathcal{A}$ is a sigma-algebra, then $\{E_n{\}}_n\in \mathcal{A}$ $⟹{\bigcap }_n{E}_n\in \mathcal{A}$

We’ll show soon that $\mathcal{M}=\{S\subset \mathbb{R}|S\text{ measurable}\}\subseteq \mathcal{P}(\mathbb{R})$ is a sigma-algebra.

Examples of $\sigma -$algebras

Stupid examples:

• $\mathcal{A}=\{\varnothing ,\mathbb{R}\}$
• $\mathcal{A}=\mathcal{P}(\mathbb{R})$ Less stupid example: $\mathcal{A}=\{E\subset \mathbb{R}|E\text{ or }{E}^c\text{ is countable}\}$
• obviously, this is closed under complements by definition
• If $\{E_n{\}}_n\subset \mathcal{A}$ with all $E_n$ countable, then the union ${\bigcup }_n{E}_n$ is a countable union of countable sets, which is countable. Thus it is in $\mathcal{A}$
• If there exists some $N_0$ such that $E_{N_0}^c$ is countable (but $E_{N_0}$ is not), then ${\left({\bigcup }_n{E}_n\right)}^c={\bigcap }_n{E}_n^c\subset E_{N_0}^c$ is an intersection of sets, one of which is countable $⟹({\bigcup }_n{E}_n{)}^c\in \mathcal{A}$

Theorem

Let $\Sigma =\{\mathcal{A}:\mathcal{A}\text{ is a }\sigma -\text{algebra containing all open subsets of }\mathbb{R}\}$

$\mathcal{P}(\mathbb{R})\in \Sigma$

Define the intersection of all such sigma-algebras as $\mathcal{B}={\bigcap }_{\mathcal{A}\in \Sigma }\mathcal{A}$ This is the smallest $\sigma -$algebra containing all open subsets of $\mathbb{R}$ and it is called the Borel $\sigma -$algebra

($\mathcal{B}\in \Sigma$ and for all $\mathcal{A}\in \Sigma$, we have $\mathcal{B}\subset \mathcal{A}$)

see borel sigma algebra

Proof

Just need to show that $\mathcal{B}$ is indeed an $\sigma -$algebra.

• Since every open subset is an element of each $\mathcal{A}\in \Sigma$, we conclude that each is then an element of $\mathcal{B}$.
• And because it is the intersection of all other $\sigma -$algebras in $\Sigma$, it must indeed be the smallest one. Suppose $E\in \mathcal{B}$.
• Then for all $\mathcal{A}\in \Sigma ,E\in \mathcal{A}$.
• So for all $\mathcal{A}$, we have $E^c\in \mathcal{A}⟹E^c\in \mathcal{B}$
• Similarly, for any $\{E_n{\}}_n$ such that $E_n\in \mathcal{B}$ for all $n$, we must have $E_n\in \mathcal{A}$
• Then (because they are sigma-algebras) we also have ${\bigcup }_n{E}_n\in \mathcal{A}$ for all $\mathcal{A}⟹{\bigcup }_n{E}_n\in \mathcal{B}$

Next time, we will see that the set of lebesgue measurable sets is a sigma-algebra and in fact contains the borel sigma algebra.

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