# Adventures in Measure Theory - 5

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

Continuing from where I left off last time,

#### 111X (e)

Let $X$ be a set, $\mathcal A$ a family of subsets of $X$, and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\mathcal A$. Suppose that $Y$ is another set and $\phi:Y\to X$ a function. Show that $\{\phi^{-1}[E]:E\in\Sigma\}$ is the $\sigma$-algebra of subsets of $Y$ generated by $\{\phi^{-1}[A]:A\in\mathcal A\}$.

I have a confession to make. This problem had been troubling me for DAYS. It all came down to proving an equality and I was only able to prove one side of it. I eventually had to look up the solution and found one here on StackExchange. Before I start the proof however, there is an interesting comment on the selected answer that I would like to talk about here:

The general trick is that if you want to show something (call it $P$) holds for all sets $A$ in a $\sigma$-algebra $\sigma(\mathcal{C})$ generated by a known collection, the obvious approach “let $A \in \sigma(\mathcal{C})$, show $A$ satisfies $P$” is often not helpful, because you don’t know what $A$ might look like. Instead, try considering the collection $\mathcal{B}$ of all sets satisfying $P$. If you can show that $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{C}$, you will be done.

One question I had immediately upon reading this was: Why will we be done if we can show that $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{C}$? Then I realised, that the $\sigma$-algebra generated by $\mathcal{C}$ is defined as the intersection of the set of all $\sigma$-algebras that contain $\mathcal{C}$, i.e.,

\[\sigma(\mathcal{C}) = \bigcap\{\Sigma :\Sigma\text{ is a }\sigma\text{-algebra and }\mathcal{C}\subseteq\Sigma \}\]which means that $\mathcal{B}$ is also a part of this intersection, and hence, every element of $\sigma(\mathcal{C})$ must satisfy $P$ because every element of $\mathcal{B}$ satisfies $P$.

I think this is a nice approach. Anyways, let’s get back to the proof,

*Proof.*

We want to show that $\{\phi^{-1}[E]:E\in\Sigma\}$ is the $\sigma$-algebra of subsets of $Y$ generated by $\{\phi^{-1}[A]:A\in\mathcal A\}$. Or in other words, $\{\phi^{-1}[E]:E\in\Sigma\}$ equals $\bigcap T$ where

First we will show that $\bigcap T\subseteq\{\phi^{-1}[E]:E\in\Sigma\}$.

- In 111X (d) we proved that $\{\phi^{-1}[E]:E\in\Sigma\}$ is a $\sigma$-algebra of subsets of $Y$ (the domain of $\phi$).
- $\mathcal A\subseteq\Sigma$

$\implies\{\phi^{-1}[A]:A\in\mathcal{A}\}\subseteq\{\phi^{-1}[E]:E\in\Sigma\}\}$

$\implies\{\phi^{-1}[E]:E\in\Sigma\}\in T$

$\implies\bigcap T\subseteq\{\phi^{-1}[E]:E\in\Sigma\}$

Now we will show that $\{\phi^{-1}[E]:E\in\Sigma\}\subseteq\bigcap T$. For this we will use the new approach we read about above.

Let $\mathcal D$ be a family of subsets of $X$ such that for every $D$ in $\mathcal D$ we have, $\phi^{-1}[D]\in\bigcap T$, i.e.,

\[\mathcal D = \{D: D\subseteq X \text{ and }\phi^{-1}[D]\in\bigcap T \}\]Now,

(i) We know that $\emptyset\subseteq X$ and $\phi^{-1}[\emptyset]=\emptyset\in\bigcap T$. Thus, $\emptyset\in\mathcal D$.

(ii) Let $D\in\mathcal D$. For every $\phi^{-1}[D]\in\bigcap T$ we have,

\[{ Y\setminus\phi^{-1}[D]\in\bigcap T \\ \implies\phi^{-1}[X\setminus D]\in\bigcap T }\]Also, it is obvious that $X\setminus D\subseteq X$. Thus, $X\setminus D\in\mathcal D\,\,\forall\,\,D\in\mathcal D$.

(iii) Let $\langle E_n \rangle_{n\in\Bbb N}$ be a sequence of sets in $\mathcal D$. Then $\langle\phi^{-1}[E_n]\rangle_{n\in\Bbb N}\in\bigcap T$. This means that,

\[{ \bigcup_{n\in\Bbb N}\phi^{-1}[E_n]\in\bigcap T \\ \implies\phi^{-1}[\bigcup_{n\in\Bbb N}E_n]\in\bigcap T }\]Also, $\bigcup_{n\in\Bbb N}E_n\subseteq X$. Thus, $\bigcup_{n\in\Bbb N}E_n\in\mathcal D$.

From (i), (ii) and (iii) it is clear that $\mathcal D$ is a $\sigma$-algebra of subsets of $X$.

(iv) From the definition of $T$,

\[{ \{\phi^{-1}[A]:A\in\mathcal A\}\subseteq\bigcap T \\ \implies\mathcal A\subseteq\mathcal D }\]Thus, $D$ is a $\sigma$-algebra of subsets of $X$ that contains $\mathcal A$. This means that

\[\Sigma\subseteq\mathcal D\]and therefore from the definition of $\mathcal D$,

\[\phi^{-1}[E]\in\bigcap T\,\,\forall\,\, E\in\Sigma \\ \implies\{\phi^{-1}[E]: E\in\Sigma \}\subseteq\bigcap T\]We have thus proved both the sides, $\bigcap T\subseteq\{\phi^{-1}[E]:E\in\Sigma\}$ and $\{\phi^{-1}[E]: E\in\Sigma \}\subseteq\bigcap T$. Therefore, $\{\phi^{-1}[E]: E\in\Sigma \}=\bigcap T$. \(\tag*{$\square$}\)

Phew, that was one troublesome problem. If you have reached the end of this post, I commend your willpower. Let me know if there are any mistakes in my solution, or you know of a more elegant / easier / better proof for this problem. Drop an email at ricekot [at] gmail [dot] com.