Measure Theory/The Measurable Sets Form a Sigma-algebra

The Measurable Sets Form a Sigma-algebra

In a previous exercise you showed that if $E\in ℳ$ then $E^c\in ℳ$. This is called "closure under complements".

Here we will prove other useful closure properties, such as closure under unions and closure under intersections.

Suppose that $ℳ$ is closed under all unions. That is to say, if $ℰ\subseteq 𝒫(ℳ)$ is any collection of measurable sets, then

$\bigcup _{E\in ℰ}E\in ℳ$

Show that, in that case, $ℳ=𝒫(ℝ)$.

Does this make you suspect that $ℳ$ is or is not closed under arbitrary unions? (We will answer this question more formally in the next lesson.)

Closure Under Union

Exercise 1. Not All Unions hints to us that $ℳ$ is not closed under so-called "arbitrary unions". But perhaps there is some weaker closure property for unions that it still satisfies?

Let's try to prove closure under just pairwise unions.

Let $E,F\in ℳ$ and we will try to show that $E\cup F\in ℳ$.

Let $A\subseteq ℝ$ and we need to show that $E\cup F$ splits A cleanly. As usual, we only need to show one direction,

${\lambda }^{*}(A)\ge {\lambda }^{*}(A\cap [E\cup F])+{\lambda }^{*}(A\setminus [E\cup F])$

because the other direction is handled automatically by subadditivity.

Because this proof can get complicated, it may help to name all the relevant components of the set $E\cup F$ in the following way.

• $E\cap F=E_{11}$
• $E\cap F^c=E_{10}$
• $E^c\cap F=E_{01}$
• $E^c\cap F^c=E_{00}$

Notice that with this naming system, $E\cup F=E_{11}\bigsqcup E_{10}\bigsqcup E_{01}$. Therefore the right-hand side of the inequality that we seek to prove is

${\lambda }^{*}(A\cap [E\cup F])+{\lambda }^{*}(A\setminus [E\cup F])$

which is now the same as

${\lambda }^{*}([A\cap E_{11}]\bigsqcup [A\cap E_{10}]\bigsqcup [A\cap E_{01}])+{\lambda }^{*}(A\cap E_{00})$

Complete the proof of closure under pairwise union by following these steps.

1. Strategically apply subadditivity.  That is to say, do not fully distribute the ${\lambda }^{*}$ to every unioned set above.  Keep inside of the ${\lambda }^{*}$ those sets with initial index 1.

2.  For the stuff left not distributed, reorganize this into ${\lambda }^{*}(A\cap E)$.

3.  Use the fact that F is measurable, applied to the set $A\cap E^c$, to "factor out" a ${\lambda }^{*}$.  Then reorganize this.

4.  Use the fact that E is measurable.

Closure under Finite Unions

Now that we have closure under pairwise unions, this generalizes easily to closure under finite unions.

This time I won't tell you exactly the theorem to state. Rather, it is your job to both formalize the theorem for finite unions, and then to prove it.

Closure under Countable Unions

Finally we show closure under countable unions.

Let $E_1,E_2,\dots \in ℳ$ be any countable collection of measurable sets, and $A\subseteq ℝ$. As always, we need

${\lambda }^{*}\left(A\right)\ge {\lambda }^{*}(A\cap \bigcup _{i=1}^{\mathrm{\infty }}{E}_i)+{\lambda }^{*}(A\setminus \bigcup _{i=1}^{\mathrm{\infty }}{E}_i)$

By the result for finite additivity, we know that for each $n\in \mathbb{N}$,

${\lambda }^{*}(A)\ge {\lambda }^{*}(A\cap \bigcup _{i=1}^n{E}_i)+{\lambda }^{*}(A\setminus \bigcup _{i=1}^n{E}_i)$

Explain why $\lambda (A\setminus \bigcup _{i=1}^n{E}_i)\le \lambda (A\setminus \bigcup _{i=1}^{\mathrm{\infty }}{E}_i)$

Prove that $\underset{n\to \mathrm{\infty }}{\lim }{\lambda }^{*}(A\cap \bigcup _{i=1}^n{E}_i)={\lambda }^{*}(A\cap \bigcup _{i=1}^{\mathrm{\infty }}{E}_i)$ by completing the following steps.

1. For a fixed n, prove ${\lambda }^{*}(A\cap \bigcup _{i=1}^n{E}_i)={\lambda }^{*}((A\cap \bigcup _{i=1}^n{E}_i)\cap E_n)+{\lambda }^{*}((A\cap \bigcup _{i=1}^n{E}_i)\setminus E_n)$.  Then simplify $(A\cap \bigcup _{i=1}^n{E}_i)\cap E_n$ and $(A\cap \bigcup _{i=1}^n{E}_i)\setminus E_n$.  

2. Use (1.) to prove by induction that ${\lambda }^{*}(A\cap \bigcup _{i=1}^n{E}_i)=\sum _{i=1}^n{\lambda }^{*}(A\cap E_i)$.  

3. Prove that $\sum _{i=1}^n\lambda (A\cap E_i)\le \sum _{i=1}^{\mathrm{\infty }}\lambda (A\cap E_i)$.

4. Take a limit as $n\to \mathrm{\infty }$ and infer the result.

Intersections

Use closure under complements and closure under countable unions to easily infer closure under countable intersections.

Sigma Algebra

Prove that $ℳ$ is a $\sigma$-algebra. (The proof should merely reference results that we've already proved elsewhere.)

(Note: As of right now the concept of a $\sigma$-algebra isn't terribly useful. However, if and when I extend this course to cover topics in general measure theory rather than just Lebesgue measure, the concept will become more important.)

All Intervals

In a previous lesson we showed that the open rays, $(a,\mathrm{\infty })\subseteq ℝ$, are all measurable.

Now use the $\sigma$-algebra properties of $ℳ$ to show that every other interval is measurable. Note: Intervals may be closed, open, or neither. They may also be bounded or unbounded.