Exercise: Let $(\Omega, \mathcal{A})$ and $(E, \mathcal{B})$ be measurable spaces, and let $(f_{n})_{n\geq 1}$ be a sequence of measurable functions from $\Omega$ to $E$. Let $t$ be a measurable function from $\Omega$ to $\mathbb{N}^{*}$. Prove that the function $\varphi$ defined by: $$ \begin{aligned} \varphi : \Omega &\longrightarrow E \\ x &\longmapsto f_{t(x)}(x) \end{aligned} $$ is measurable.
Solution: Let $B$ be an element of the $\sigma$-algebra $\mathcal{B}$. We have: $$ \varphi^{-1}(B) = \bigcup_{n\geq 1} \left\{ \{t=n\} \cap f_{n}^{-1}(B) \right\} \in \mathcal{A} $$ Therefore, $\varphi$ is a measurable function.
Could someone kindly explain the meaning and steps involved in the equation
$$\varphi^{-1}(B) = \bigcup_{n\geq 1} \left\{ \{t=n\} \cap f_{n}^{-1}(B) \right\} \in \mathcal{A}$$
I'm studying measure theory and would appreciate some clarification on this equality and its implications within this context.
Thank you for your assistance!
Just start from the definition of the preimage : according to the definition, given a subset $B\subseteq E$, $\varphi^{-1}(B)\subseteq \Omega $ is the set of all the $x$ in $\Omega$ such that $\varphi(x)\in B$.
So writing this more formally and replacing $\varphi$ with its definition this becomes $$\varphi^{-1}(B) := \{x\in\Omega:f_{t(x)}(x) \in B\} $$ Now we need to unpack this definition a bit more : for a fixed $x\in\Omega$, what does $f_{t(x)}(x)\in B $ mean ?
Well, first note that $t$ takes values in $\mathbb N^*$, so there exists $n\in\mathbb N^*$ such that $t(x) = n$. So for this fixed $x$, we see that $f_{t(x)}(x)\in B $ if and only if $f_n(x)\in B $. But again, by definition of the preimage, $f_n(x)\in B$ if and only if $x\in f_n^{-1}(B)$. So what this proves is that $$ \varphi^{-1}(B) := \{x\in\Omega:t(x) = n\text{ for some $n\in\mathbb N^*$, and } x \in f_n^{-1}(B)\} $$
But now we are almost done : indeed the set $\{x\in\Omega : \exists n\in\mathbb N^* : t(x) = n\} $ is exactly equal to $\bigcup_{n\in\mathbb N^*}\{x\in\Omega : t(x) = n\} $ (if you're not convinced, it's a good exercise to prove it), while the set $\{x\in\Omega :x\in f_n^{-1}(B)\}$ is by definition $f_n^{-1}(B)$. Since logical "and" translates to set intersection, it follows that indeed $$\varphi^{-1}(B) :=\bigcup_{n\in\mathbb N^*}\left\{\{x\in\Omega : t(x) = n\}\cap f_n^{-1}(B)\right\} $$ And in most cases we use the shorter notation $\{t = n\} $ for $\{x\in\Omega : t(x) = n\} $ if there is no ambiguity.
Lastly, because $f_n$ is measurable for all $n$, we have that $f_n^{-1}(B)\in\mathcal{A}$, while $\{t = n\} $ (which can also be written $t^{-1}(\{n\})$) is also in $\mathcal{A}$ for the same reason. Hence $\{t = n\} \cap f_n^{-1}(B)\in \mathcal{A}$ and so we have that $\varphi^{-1}(B)$, as a countable union of elements of $\mathcal{A}$, is also in $\mathcal{A}$. This shows that the preimage of any measurable set under $\varphi$ is measurable, which means by definition that $\varphi$ is measurable.
Does that help ?