# Adventures in Measure Theory - 3

I am following the Measure Theory series by D.H. Fremlin and blogging my notes here.

Today we’re going to solve some problems!

#### 111X (a)

The nine pairs are:

- $E\cap(\bigcup_{n\in\Bbb N}F_n) = \bigcup_{n\in\Bbb N}(E\cap F_n)$
- $E\cup(\bigcap_{n\in\Bbb N}F_n) = \bigcap_{n\in\Bbb N}(E\cup F_n)$
- $\bigcup_{n\in\Bbb N}(E_n\setminus F) = (\bigcup_{n\in\Bbb N}E_n)\setminus F$
- $E\setminus(\bigcup_{n\in\Bbb N}F_n) = \bigcap_{n\in\Bbb N}(E\setminus F_n)$
- $\bigcup_{n\in\Bbb N}(E\setminus F_n) = E\setminus(\bigcap_{n\in\Bbb N}F_n)$
- $\bigcap_{n\in\Bbb N}(E_n\setminus F) = (\bigcap_{n\in\Bbb N}E_n)\setminus F$
- $\bigcap_{m,n\in\Bbb N}(E_m\setminus F_n) = (\bigcap_{n\in\Bbb N}E_n)\setminus(\bigcup_{n\in\Bbb N}F_n)$
- $(\bigcup_{n\in\Bbb N}E_n)\cap(\bigcup_{n\in\Bbb N}F_n) = \bigcup_{m,n\in\Bbb N}(E_m\cap F_n)$
- $(\bigcap_{n\in\Bbb N}E_n)\cup(\bigcap_{n\in\Bbb N}F_n) = \bigcap_{m,n\in\Bbb N}(E_m\cup F_n)$

#### 111X (b)

##### All ‘open intervals’ $(a,b)$, $(-\infty, b)$, $(a, \infty)$ are open sets.

*Proof.*

Consider an interval $(a,b)$. We want to show that $\forall\,\, x\in (a,b)\,\,\exists\,\,\delta > 0$ such that $(x-\delta,x+\delta)\in (a,b)$. If we set $\delta = \min\{x-a, b-x\}$ we are done.

- How did we get this value of $\delta$?

Since $\delta$ will be different for every $x$, we can deduce that it must be a function of $x$. Now we want a value of $\delta$ such that $x - \delta \ge a$ and $x + \delta \le b$. The largest permissible value of $\delta$ will be the minimum of the two values we get when these two inequalities are written as equalities i.e. either $x - \delta = a$ or $x + \delta = b$.

##### All intervals (bounded or unbounded, open, closed or half-open) are Borel sets.

*Proof.*

Borel sets are the members of the $\sigma$-algebra of subsets of $\Bbb R$ generated by the open sets of $\Bbb R$. We just proved above that all open intervals (bounded and unbounded) are open sets, and hence are also Borel sets. Now we have to prove that closed and half-open intervals are also Borel sets.

For any open interval $(a, b)$ in a Borel $\sigma$-algebra, it’s complement $(-\infty,a]\cup[b, \infty)$ also lies in the Borel $\sigma$-algebra. Likewise we can construct all possible half-open or closed intervals simply by taking the union, complement or perhaps both of the Borel sets we already have (or get).

There must be a more elegant proof; I am not convinced with the two paragraphs I have whipped up. Based on what I found here on StackExchange, here is a much better proof:

The initial argument remains the same. All open intervals must be Borel sets (by definition). We will now show that closed and half-open intervals can be expressed as open intervals.

Consider the sets $[a, b]$ and $\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})$. It is clear that $[a, b] \subseteq (a-\frac{1}{n}, b+\frac{1}{n})\forall n\in\Bbb N$. Thus,

\[[a, b] \subseteq\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n}) \tag{i}\]Now, let $x\in\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})$. We will prove that $x\in [a, b]$ by contradiction. Let us assume that $x\notin [a, b]$ or more specifically, $x < a$. Then, by the Archimedean property, we can say that $\exists\,\, n\in\Bbb N$ such that $1 < n(a-x)$ or $x < a - \frac{1}{n}$. Similarly, if $x > b$ then $x > b+\frac{1}{n}$ for some $n\in\Bbb N$. These two inequalities are a contradiction because $a-\frac{1}{n} < x < b+\frac{1}{n} \,\,\forall\,\, n\in\Bbb N$. This means that our assumption was wrong and $x\in [a, b]$ indeed. Thus,

\[\bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})\subseteq [a, b]\tag{ii}\]From $(i)$ and $(ii)$, we have,

\[[a, b] = \bigcap_{n\in\Bbb N}(a-\frac{1}{n}, b+\frac{1}{n})\]This means that $[a, b]$ is also an open interval and hence a Borel set.

In a similar fashion we can prove that,

\[(a, b] = \bigcap_{n\in\Bbb N} (a, b + \frac{1}{n}) \\ [a, b) = \bigcap_{n\in\Bbb N} (a - \frac{1}{n}, b)\]Hence, proved.