Functional Analysis Lecture 7

[[lecture-data]]

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

2. Lebesgue Measure and Integration

Recall we defined

Outer Measure

For $A \subset R$ , we define the outer measure of $A$
$m^{*} (A) = inf {\sum_{n} ℓ (I_{n}) : {I_{n}} countable and open s.t. A \subset ⋃_{n} I_{n}}$
We can see that $m^{*} (A) \geq 0$ for all $A$ .

Theorem

If $I$ is an interval, then

m^{*} (I) = ℓ (I)

Proof

Suppose $I = [a, b]$ . Then $I \subset (a - ϵ, b + ϵ)$ for all $ϵ > 0$ . But then we have

m^{*} (I) \leq ℓ (a - ϵ, b + ϵ) = b - a + 2 ϵ

Thus we have $m^{*} (I) \leq b - a$ .

Now we show $b - a \leq m^{*} (I)$ . Let ${I_{n}}_{n}$ be a collection of open intervals such that $[a, b] \subset ⋃_{n} I_{n}$ .

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 $[a, b] \subset ⋃_{k = 0}^{m} J_{k}$

Since $a \in ⋃_{k = 0}^{m} J_{k}$ , 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 $K, 1 \leq K \leq m$ such that $b < b_{K}$ . And for all $k \in {1, 2, \dots, K - 1}$ we have $b_{k} \leq b$ and $a_{k + 1} \leq b_{k} < b_{k + 1}$ .

But then we have

\begin{aligned} \sum_{n} ℓ (I_{n}) & \geq \sum_{k = 1}^{m} ℓ (J_{k}) \\ \geq \sum_{k = 1}^{K} ℓ (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 \\ = ℓ (I) \end{aligned}

Thus we have $m^{*} (I) \geq ℓ (I) = b - a$

Thus we conclude $m^{*} (I) = ℓ (I)$

\begin{matrix} ◼ \end{matrix}

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

\begin{aligned} [a + ϵ, b - ϵ] & \subset I & \subset [a - ϵ, b + ϵ] \\ ⟹ m^{*} ([a + ϵ, b - ϵ]) & \leq m^{*} (I) & \leq m^{*} ([a - ϵ, b + ϵ]) \\ ⟹ (b - a) - 2 ϵ & \leq m^{*} (I) & \leq (b - a) + 2 ϵ \end{aligned}

Taking $ϵ \to 0$ , we get the desired result.

Exercise

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

Theorem

for every $A \subset R$ and $ϵ > 0$ , there exist an open set $O$ such that $A \subset O$ and

m^{*} (A) \leq m^{*} (O) \leq m^{*} (A) + ϵ

Proof

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

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

\begin{aligned} m^{*} (O) & = m^{*} (⋃_{n} I_{n}) \\ \leq \sum_{n} m^{*} (I_{n}) \\ \leq \sum_{n} ℓ (I_{n}) \\ \leq m^{*} (A) + ϵ \end{aligned}

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

Measurable Sets

Lebesgue measurable

A set $E \subset R$ is Lebesgue measurable if for all $A \subset R$ we have

m^{*} (A) = m^{*} (A \cap E) + m^{*} (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 $m^{*} (A) \leq m^{*} (A \cap E) + m^{*} (A \cap E^{c})$ .

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

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

[[lebesgue measurable]]

Theorem

The empty set $\emptyset$ and the set of real numbers $R$ 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 $m^{*} (E) = 0$ then $E$ is measurable.

Proof

Let $A \subset R$ . Then $A \cap E \subset E$ means

m^{*} (A \cap E) \leq m^{*} (E) = 0 ⟹ m^{*} (A \cap E) = 0

But then we see

m^{*} (A \cap E) + m^{*} (A \cap E^{c}) = m^{*} (A \cap E^{c}) \leq m^{*} (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 $A \subset R$ . Since $E_{2}$ is measurable, we know

m^{*} (A \cap E_{1}^{c}) = m^{*} (A \cap E_{1}^{c} \cap E_{2}) + m^{*} (A \cap \underset{(E_{1} \cup E_{2})^{c}}{\underset{⏟}{E_{1}^{c} \cap E_{2}^{c}}})

Then we also know

\begin{aligned} 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{aligned}

Taking outer measure we get

\begin{aligned} ⟹ 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} measurable ⟹ & = 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{aligned}

And this gives us the desired result.

see [[unions of measurable sets are measurable]]

Theorem

If $E_{1}, \dots, E_{n}$ are measurable, then $⋃_{k = 1}^{n} E_{k}$ is measurable

Proof

by induction.
$n = 1$ is trivial. Suppose the statement holds for $n = k$ . Then

⋃_{j = 1}^{k + 1} E_{j} = (⋃_{j = 1}^{k} E_{j}) \cup E_{k + 1}

And then since [[unions of measurable sets are measurable]] we get the desired result.

Algebra

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

(closure under complements) $E \in A ⟹ E^{c} \in A$
(closure under finite unions) $E_{1}, \dots, E_{n} \in A ⟹ ⋃_{k = 1}^{n} E_{k} \in A$

see [[algebra]], [[sigma-algebra]]

σ -

algebra

We say an [[algebra]] is a $σ -$ algebra (sigma-algebra) if also
3. (closure under countable unions) ${E_{n}}_{n = 1}^{\infty} \in A ⟹ ⋃_{n} E_{n} \in A$

Note

By DeMorgan's Laws, if $E_{1}, \dots, E_{n} \in A$ , then we also have

⋂_{k = 1}^{n} E_{k} = {(⋃_{k = 1}^{n} E_{n})}^{c} \in A

Thus, if $E \in A$ then

\emptyset = E \cap E^{c} \in A ⟹ R = \emptyset^{c} \in A

Similarly, if $A$ is a [[sigma-algebra]], then ${E_{n}}_{n} \in A$

⟹ ⋂_{n} E_{n} \in A

We'll show soon that

M = {S \subset R | S measurable} \subseteq P (R)

is a [[sigma-algebra]].

Examples of

σ -

algebras

Stupid examples:

$A = {\emptyset, R}$
$A = P (R)$
Less stupid example:

A = {E \subset R | E or E^{c} is countable}

obviously, this is closed under complements by definition
If ${E_{n}}_{n} \subset A$ with all $E_{n}$ countable, then the union $⋃_{n} E_{n}$ is a countable union of countable sets, which is countable. Thus it is in $A$
If there exists some $N_{0}$ such that $E_{N_{0}}^{c}$ is countable (but $E_{N_{0}}$ is not), then ${(⋃_{n} E_{n})}^{c} = ⋂_{n} E_{n}^{c} \subset E_{N_{0}}^{c}$ is an intersection of sets, one of which is countable $⟹ (⋃_{n} E_{n})^{c} \in A$

Theorem

Let

Σ = {A : A is a σ - algebra containing all open subsets of R}

Example

$P (R) \in Σ$

Define the intersection of all such sigma-algebras as

B = ⋂_{A \in Σ} A

This is the smallest $σ -$ algebra containing all open subsets of $R$ and it is called the Borel $σ -$ algebra

( $B \in Σ$ and for all $A \in Σ$ , we have $B \subset A$ )

see [[borel sigma algebra]]

Proof

Just need to show that $B$ is indeed an $σ -$ algebra.

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

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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