# A function $f: \mathbb{N} \times Y \to \mathbb{R}$ is measurable iff each section $f_n$ is $\mathcal{Y}$-measurable.

Let $(\mathbb{N}, \mathcal{P}\mathbb{N}, \mu)$ be the measure space on the natural numbers with $\mu$ as the counting measure. Let $(Y, \mathcal{Y}, \nu)$ be an arbitrary measure space.

I want to show that $f: \mathbb{N} \times Y \to \mathbb{R}$ is measurable iff each section $f_n$ is $\mathcal{Y}$-measurable. In this context, a section $f_n: Y \to \mathbb{R}$ is defined via $$f_n(y) = f(n,y), \quad y \in Y$$ where $f: \mathbb{N} \times Y \to \mathbb{R}$.

Thoughts:

Given that $f$ is a measurable function, there's a theorem that allows me to conclude that every section of $f$ is measurable so the forward direction is done. For the backward direction, assuming each $f_n$ is $\mathcal{Y}$-measurable I have that $$f_n^{-1}(\alpha, \infty] \in \mathcal{Y}$$ for each $n \in \mathbb{N}$ and want to show this yields for $$f^{-1}(\alpha, \infty] \in \mathcal{P}\mathbb{N} \times \mathcal{Y}.$$

I know that I can write $f^{-1}(\alpha, \infty] = \{(n,y): f(n,y) > \alpha\}.$

Any ideas on how to prove this direction? General tips/strategies would be much appreciated.

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Hint: $f^{-1}((\alpha,\infty])=\bigcup_n \left(\{n\}\times f_n^{-1}((\alpha,\infty])\right)$ –  Julian Rosen Dec 3 '12 at 4:01
Not an official hw problem, though it is taken from Bartle's Elements of integration.. –  tentaclenorm Dec 3 '12 at 4:01
@PinkElephants Could you elaborate on this fabulous hint? It was something I thought would be true, but I'm just not seeing it.. –  tentaclenorm Dec 3 '12 at 4:08

If $S\subset Y$ is measureable, then $\{n\}\times S\subset \mathbb{N}\times Y$ is measurable for all $n\in\mathbb{N}$ (it is a product of the measurable sets $\{n\}\subset \mathbb{N}$ and $S\subset Y$). Therefore $$f^{-1}((\alpha,\infty])=\bigcup_n \{n\}\times f_n^{-1}((\alpha,\infty])$$ is a countable union of measurable sets, so is measurable.
I still don't see why $f^{-1}((\alpha,\infty])=\bigcup_n \{n\}\times f_n^{-1}((\alpha,\infty])$ is true. Is there a way to achieve it starting from $f^{-1}((\alpha,\infty])=\{(n,y): f_n(y) > \alpha\}$ ? –  tentaclenorm Dec 3 '12 at 4:20
Take the set $A:=\{(n,y):n\in\mathbb{N},f_n(y)>y\}$, and decompose it as a union $A=\bigcup_k A_k$ of subsets $A_k\subset A$, where $A_k=\{(n,y)\in A:n=k\}$. Do you see why $A_k=\{k\}\times f_k^{-1}((\alpha,\infty])$? –  Julian Rosen Dec 3 '12 at 4:27