Functional Analysis Lecture 8

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Lecture Notes: Rodriguez, page 36

2. Lebesgue measure and integration

Recall from last time

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) $E_{1}, \dots, E_{n} \in A ⟹ ⋃_{k = 1}^{n} E_{k} \in A$

σ -

algebra

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

Lebesgue Measurable

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

Today we will

show that the borel sigma algebra is a subset of all measurable sets $M$
define Lebesgue measure finally

Lemma

Let $A$ be an algebra and let ${E_{n}}$ be a countable collection of elements in $A$ . Then there exists ${F_{n}}$ a disjoint countable collection of elements in $A$ such that

⋃_{n} E_{n} = ⋃_{n} F_{n}

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

Proof

Let $G_{n} = ⋃_{k = 1}^{n} E_{k}$ so that $G_{1} \subset G_{2} \subset G_{3} \subset \dots$ and $⋃_{n} G_{n} = ⋃_{n} E_{n}$ . Now, let $F_{1} = G_{1}$ and for all $k$ define

F_{k + 1} = G_{k + 1} ∖ G_{k}

Then we also have that the disjoint $⋃_{n} F_{n} = ⋃_{n} G_{n} = ⋃_{n} E_{n}$ .

\begin{matrix} ◼ \end{matrix}

see algebras have closure under finite disjoint countable unions

Theorem

Let $A \subset R$ and let $E_{1}, \dots, E_{n}$ be disjoint and measurable. Then

m^{*} (A \cap [⋃_{k = 1}^{n} E_{k}]) = \sum_{k = 1}^{n} m^{*} (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 $A \subset R$ . Since $E_{k} \cap E_{m + 1} = \emptyset$ for all $k$ , we have (since $E_{m + 1}$ is measurable ) that

\begin{aligned} A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1} & = A \cap E_{m + 1} \\ and A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}^{c} & = A \cap [⋃_{k = 1}^{m} E_{k}] \\ ⟹ m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}]) & = m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}) + m^{*} (A \cap [⋃_{k = 1}^{m + 1} E_{k}] \cap E_{m + 1}^{c}) \\ = m^{*} (A \cap E_{m + 1}) + m^{*} (A \cap [⋃_{k = 1}^{m} E_{k}]) \\ = m^{*} (A \cap E_{m + 1}) + \sum_{k = 1}^{m} m^{*} (A \cap E_{k}) \end{aligned}

Where the last inequality is due to the induction hypothesis. Thus the result follows via induction.

\begin{matrix} ◼ \end{matrix}

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

Theorem

The collection of measurable sets $M$ is a sigma-algebra

Proof

We already know that $M$ 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 $E = ⋃_{n} E_{n}$ . Then since the other half is always satisfied via monotonicity, we just need to show $m^{*} (A \cap E^{c}) + m^{*} (A \cap E) \leq m^{*} (A)$

Let $N \in N$ . Since $M$ is an algebra, the union $⋃_{n = 1}^{N} E_{n} \in M$ so we have

m^{*} (A) = m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap {[⋃_{n = 1}^{N} E_{n}]}^{c})

And since $⋃_{n = 1}^{N} E_{n} \subset E$ , we have $E^{c} \subset {[⋃_{n = 1}^{N} E_{n}]}^{c}$ . Thus we have

\begin{aligned} m^{*} (A) & = m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap {[⋃_{n = 1}^{N} E_{n}]}^{c}) \\ \geq m^{*} (A \cap [⋃_{n = 1}^{N} E_{n}]) + m^{*} (A \cap E^{c}) \end{aligned}

And since the measure of finite disjoint measurable sets is the sum of the measures, we get

\begin{aligned} 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^{*} (⋃_{n} A \cap E_{n}) + m^{*} (A \cap E^{c}) \\ = m^{*} (A \cap E) + m^{*} (A \cap E^{c}) \end{aligned}

Where $(*)$ is from countable subadditivity.

\begin{matrix} ◼ \end{matrix}

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

$m (E)$ is defined for all $E \subset R$
if $I$ is an interval then $m (E) = ℓ (I)$ the "length" of $I$
If ${E_{n}}$ is a (countable) collection of disjoint subsets of $E$ such that $E = ⋃_{n} E_{n}$ , then we want $m (⋃_{n} E_{n}) = m (E)$
Translation invariance. ie, if $E \subset R$ and $x \in R$ , 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 $B$ the borel sigma algebra. To do this, we just need to show that it contains all open sets.

Proposition

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

Proof

Suppose $A \subset R$ . Let $A_{1} = A \cap (a, \infty)$ and $A_{2} = A \cap (- \infty, a]$ . Now we want to show that $m^{*} (A_{1}) + m^{*} (A_{2}) \leq m^{*} (A)$ .

If $m^{*} (A)$ is infinite we are done. So suppose that $m^{*} (A) < \infty$ . Now, let ${I_{n}}$ be a collection of intervals such that

\sum_{n} ℓ (I_{n}) \leq m^{*} (A) + ε

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

$A_{1} \subset ⋃_{n} J_{n}$
$A_{2} \subset ⋃_{n} K_{n}$
$ℓ (I_{n}) = ℓ (J_{n}) + ℓ (K_{n})$
Thus we have

\begin{aligned} m^{*} (A_{1}) + m^{*} (A_{2}) & \leq \sum_{n} m^{*} (J_{n}) + m^{*} (K_{n}) \\ = \sum_{n} ℓ (J_{n}) + ℓ (K_{n}) \\ = \sum_{n} ℓ (I_{n}) \\ \leq m^{*} (A) + ε \end{aligned}

Then take $ε \to 0$ and we have the desired result.

\begin{matrix} ◼ \end{matrix}

see open intervals with upper bound infinity are measurable

Theorem

Every open subset of $R$ is measurable (ie $B \subset M$ - 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 $a \in R$ (open intervals with upper bound infinity are measurable), so is

(- \infty, b) = ⋃_{n = 1}^{\infty} (- \infty, b - \frac{1}{n}] = ⋃_{n = 1}^{\infty} {(b - \frac{1}{n}, \infty)}^{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 $R$ is a countable union of open intervals. Thus all open intervals are measurable.

\begin{matrix} ◼ \end{matrix}

see all borel sets are measurable

Lebesgue Measure

The Lebesgue measure of a measurable set $E \subset M$ is given by

m (E) = m^{*} (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 (⋃_{n} E_{n}) = \sum_{n} m (E_{n})

Note

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

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 (⋃_{n} E_{n}) = m^{*} (⋃_{n} E_{n}) \leq \sum_{n} m^{*} (E_{n}) = \sum_{n} m (E_{n})

So we just need to show the other way. For any $N \in N$ , since measure of finite disjoint measurable sets is the sum of the measures, we have

\begin{aligned} m (⋃_{n = 1}^{N} E_{n}) & = m^{*} (R \cap ⋃_{n = 1}^{N} E_{n}) \\ = \sum_{n = 1}^{N} m^{*} (R \cap E_{n}) \\ = \sum_{n = 1}^{N} m^{*} (E_{n}) \\ = \sum_{n = 1}^{N} m (E_{n}) \\ (*) & = m (⋃_{n = 1}^{N} E_{n}) \\ \leq m (⋃_{n = 1}^{\infty} E_{n}) \end{aligned}

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.

\begin{matrix} ◼ \end{matrix}

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 $E \in M$ and $x \in R$ 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 (⋃_{k = 1}^{\infty} E_{k}) = lim_{n \to \infty} m (⋃_{k = 1}^{n}) = lim_{n \to \infty} m (E_{n})

Proof

The second equality is because $E_{n} = ⋃_{k = 1}^{n} E_{k}$ . So it is enough to just show $m (⋃_{k} E_{k}) = lim_{n \to \infty} m (E_{n})$ .

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 $F_{k} = E_{k} ∖ E_{k - 1}$ . 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 $n \in N$ , we have

\begin{aligned} ⋃_{k = 1}^{n} F_{k} = E_{n} & ⋃_{k = 1}^{\infty} F_{k} = ⋃_{k = 1}^{\infty} E_{k} \\ ⟹ m (⋃_{k = 1}^{\infty} E_{k}) & = \sum_{k = 1}^{\infty} m (F_{k}) \\ = lim_{n \to \infty} \sum_{k = 1}^{n} m (F_{k}) \\ = lim_{n \to \infty} m (⋃_{k = 1}^{n} F_{k}) \\ = lim_{n \to \infty} m (E_{n}) \end{aligned}

Since measure of finite disjoint measurable sets is the sum of the measures. Thus we have shown the desired equality.

\begin{matrix} ◼ \end{matrix}

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.

Created 2025-06-24 Last Modified 2025-07-14