# Proving a sufficient and necessary condition for $f:\, X\to\mathbb{R}\cup\{\pm\infty\}$ to be measurable

I saw the following question:

Denote $\overline{\mathbb{R}}=\mathbb{R}\cup\{\pm\infty\}$, the open sets containing $x\in\mathbb{R}$ are the open sets in $\mathbb{R}$ containing $x$. The open sets containing $\pm\infty$ are the sets of the form $V\cup\{\infty\}$ or $V\cup\{-\infty\}$ accordingly.

Let $(X,S)$ be a measurable space. Prove that $f:\, X\to\overline{\mathbb{R}}$ is measurable iff the following conditions hold:

1. $f^{-1}(\{\infty\}),f^{-1}(\{-\infty\})\in S$

2. $f$ is measurable as a function $f^{-1}(\mathbb{R})\to\mathbb{R}$

My work:

I started with assuming that both conditions hold and I took some $B\in\mathcal{B}(\overline{\mathbb{R}})$ and I wanted to prove that $f^{-1}(B)\in S$, but the problem I have here that I don't really understand how those sets $B\in\mathcal{B}(\overline{\mathbb{R}})$ look like.

So I thought that if I will understand what $S'$ generates $\mathcal{B}(\overline{\mathbb{R}})$ I could use that. I know that the open rays of the form $(-\infty,a)$ or $(a,\infty)$ generate $\mathcal{B}(\mathbb{R})$, and it seems that the sets of the form $(-\infty,a)\cup\{\infty\}$ plus the sets of the form $(a,\infty)\cup\{\infty\}$.

Since $$f^{-1}((-\infty,a)\cup\{\infty\})=f^{-1}((-\infty,a))\cup f^{-1}(\{\infty\})$$ and I know that $f^{-1}(\{\infty\})\in S$ and since its a $\sigma-$algebra I need to prove that $f^{-1}((-\infty,a))\in S$.

I'm guessing this follows from the second condition somehow, but I don't understand how being measurable in $\mathcal{B}(\overline{\mathbb{R}})$ have anything to do with being in $S$ which is a collection of subsets of some space $X$.

Can someone please help me with this question ? am I even on the right track ?

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What do you mean by: $f$ is measurable as a function $f^{−1}(\mathbb{R})\to \mathbb{R}$? I think the logical condition would be that for any measurable set $A$ in $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, $f^{-1}(A)\in S$, which would help you out immediately... – Jan Keersmaekers Dec 11 '12 at 10:20
@JanKeersmaekers - please see the answer here: math.stackexchange.com/questions/236788/… – Belgi Dec 11 '12 at 10:26
@JanKeersmaekers - I understand it as $f^{-1}(S)\in\mathcal{B}(Y)$ where $S\in\mathcal{B}(\mathbb{R}),Y=f^{-1}(\mathbb{R})$ – Belgi Dec 11 '12 at 10:28

## 1 Answer

I assume that you are topologising $\overline{\mathbb{R}}$ so that the basic open neighbourhoods of $+ \infty$ are of the form $( a , + \infty ] = ( a , + \infty ) \cup \{ + \infty \}$ for $a \in \mathbb{R}$ (and similarly for the basic open neighbourhoods of $- \infty$).

A couple observations that should help you:

• By this previous question it turns out that the Borel subsets of $\overline{\mathbb{R}}$ are precisely those sets of the form $B \cup F$ where $B \subseteq \mathbb{R}$ is Borel, and $F \subseteq \{ - \infty , + \infty \}$.

It thus follows that $\{ - \infty \} , \{ + \infty \}$ are Borel subsets of $\overline{\mathbb{R}}$, and also every Borel subset of $\mathbb{R}$ is a Borel subset of $\overline{\mathbb{R}}$.

• Given $Y \subseteq X$, let $S^\prime = \{ A \cap Y : A \in S \}$ denote the restriction of the $\sigma$-algebra $S$ to $Y$. If $Y \in S$ it follows that $S^\prime = \{ A \in S : A \subseteq Y \}$.

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