# Measure Theory/The Measurable Sets Form a Sigma-algebra
## The Measurable Sets Form a Sigma-algebra

In a previous exercise you showed that if $E\in {\mathcal {M}}$ then $E^{c}\in {\mathcal {M}}$. This is called "closure under complements".
Here we will prove other useful closure properties, such as closure under unions and closure under intersections.
* Exercise 1. Not All Unions
* Suppose that ${\mathcal {M}}$ is closed under all unions. That is to say, if ${\mathcal {E}}\subseteq {\mathcal {P}}({\mathcal {M}})$ is any collection of measurable sets, then
* $\bigcup _{E\in {\mathcal {E}}}E\in {\mathcal {M}}$
* Show that, in that case, ${\mathcal {M}}={\mathcal {P}}({\mathbb {R}})$.
* Does this make you suspect that ${\mathcal {M}}$ 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 ${\mathcal {M}}$ 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 $E,F\in {\mathcal {M}}$ and we will try to show that $E\cup F\in {\mathcal {M}}$.
Let $A\subseteq {\mathbb {R}}$ and we need to show that $E\cup F$ splits A cleanly. As usual, we only need to show one direction,
    $\lambda ^{*}(A)\geq \lambda ^{*}(A\cap [E\cup F])+\lambda ^{*}(A\smallsetminus [E\cup F])$
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, $E\cup F=E_{11}\sqcup E_{10}\sqcup E_{01}$. Therefore the right-hand side of the inequality that we seek to prove is
    $\lambda ^{*}(A\cap [E\cup F])+\lambda ^{*}(A\smallsetminus [E\cup F])$
which is now the same as
    $\lambda ^{*}([A\cap E_{11}]\sqcup [A\cap E_{10}]\sqcup [A\cap E_{01}])+\lambda ^{*}(A\cap E_{00})$
* Exercise 2. Strategic Subadditivity
* 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.
* Exercise 3. Formalize 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 $E_{1},E_{2},\dots \in {\mathcal {M}}$ be any countable collection of measurable sets, and $A\subseteq {\mathbb {R}}$. As always, we need
    $\lambda ^{*}\left(A\right)\geq \lambda ^{*}\left(A\cap \bigcup _{i=1}^{\infty }E_{i}\right)+\lambda ^{*}\left(A\smallsetminus \bigcup _{i=1}^{\infty }E_{i}\right)$
By the result for finite additivity, we know that for each $n\in {\mathbb {N}}$,
    $\lambda ^{*}(A)\geq \lambda ^{*}\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)+\lambda ^{*}\left(A\smallsetminus \bigcup _{i=1}^{n}E_{i}\right)$
* Exercise 4. Part of Countable Unions
* Explain why $\lambda \left(A\smallsetminus \bigcup _{i=1}^{n}E_{i}\right)\leq \lambda \left(A\smallsetminus \bigcup _{i=1}^{\infty }E_{i}\right)$
* Exercise 5. The Other Part of Countable Unions
* Prove that $\lim _{n\to \infty }\lambda ^{*}\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)=\lambda ^{*}\left(A\cap \bigcup _{i=1}^{\infty }E_{i}\right)$ by completing the following steps.
* 1. For a fixed n, prove $\lambda ^{*}\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)=\lambda ^{*}\left(\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)\cap E_{n}\right)+\lambda ^{*}\left(\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)\smallsetminus E_{n}\right)$. Then simplify $\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)\cap E_{n}$ and $\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)\smallsetminus E_{n}$.
* 2. Use (1.) to prove by induction that $\lambda ^{*}\left(A\cap \bigcup _{i=1}^{n}E_{i}\right)=\sum _{i=1}^{n}\lambda ^{*}(A\cap E_{i})$.
* 3. Prove that $\sum _{i=1}^{n}\lambda (A\cap E_{i})\leq \sum _{i=1}^{\infty }\lambda (A\cap E_{i})$.
* 4. Take a limit as $n\to \infty$ and infer the result.

### Intersections

* Exercise 6. Closed under Countable Intersection
* Use closure under complements and closure under countable unions to easily infer closure under countable intersections.

### Sigma Algebra

Definition: $\sigma$-algebra
Let X be any set (of real numbers, or anything else) and let ${\mathcal {S}}\subseteq {\mathcal {P}}(X)$. If ${\mathcal {S}}$ satisfies the following three properties then we call ${\mathcal {S}}$ a $\sigma$-algebra.
* $\emptyset \in {\mathcal {S}}$
* closure under complements
* closure under countable unions
Note that the symbol $\sigma$, when transliterated into English, is written "sigma".
* Exercise 7. Measurable Sets Form a Sigma-algebra
* Prove that ${\mathcal {M}}$ 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, $(a,\infty )\subseteq {\mathbb {R}}$, are all measurable.
* Exercise 8. All Intervals Are Measurable
* Now use the $\sigma$-algebra properties of ${\mathcal {M}}$ to show that every other interval is measurable. Note: Intervals may be closed, open, or neither. They may also be bounded or unbounded.