# A question on Borel measurability

Let $$(X, \mathcal{B}_{X}, \mu)$$ be a measure space. Here, $$\mu$$ is an infinite Borel measure and $$\mu$$ is not $$\sigma$$-finite. Let $$\pi$$ be surjective Borel measurable map form $$(X, \mathcal{B}_{X}, \mu)$$ to a standard Borel space $$(Y, \mathcal{B}_{Y})$$.

My question is that whether the map: $$\tau: (Y, \mathcal{B}_{Y}) \mapsto (\overline{\mathbb{R}}, \mathcal{B}_{\overline{\mathbb{R}}})$$

$$y\longrightarrow \mu(\{x\in X: \pi(x)=y \})$$ is Borel measurable or not? Here $$\overline{\mathbb{R}}=\mathbb{R}\cup \{+\infty\}$$.

I guess the answer should be positive.

Edit: Thanks to the comments and I have modified $$Y$$ to be standard Borel.

• Without further conditions, aren't there easy counterexamples obtained by equipping $Y = \{0,1\}$ with the indiscrete $\sigma$-algebra $\mathcal{B}_Y = \{\emptyset, Y\}$? Like this, every surjective $\pi$ is measurable, but $\tau$ is not unless it's constant. Commented Jul 10 at 13:06
• @TobiasFritz $\tau$ may not even be well-defined in that case, since there’s no reason $\{x \in X: \pi(x) = y\}$ is measurable. Commented Jul 10 at 17:12

As Tobias mentioned in comments, without further assumptions you can simply equip $$Y$$ with the indiscrete $$\sigma$$-algebra. In that case $$\tau$$ may not even be well-defined, since there’s no reason $$\{x \in X: \pi(x) = y\}$$ would be measurable. Thus, I’m assuming $$\mathcal{B}_Y$$ at least contains all singletons in $$Y$$, so $$\tau$$ is ensured to be well-defined. In that case this is still false. Here is a counterexample: Let $$X = Y = [0, 1]$$, $$\mathcal{B}_X = \mathcal{B}_Y$$ be the Borel $$\sigma$$-algebra, $$A \subset [0, 1]$$ be a non-Borel subset. Let $$\mu$$ be defined by,
$$\mu(B) = \begin{cases} \infty &, \text{if }B \cap A \neq \varnothing\\ m(B) &, \text{otherwise} \end{cases}$$
where $$m$$ is the Lebesgue measure. Then $$\mu$$ is a non-$$\sigma$$-finite measure on $$([0, 1], \mathcal{B}_{[0, 1]})$$. Let $$\pi$$ simply be the identity map. Then $$\tau(y) = + \infty$$ if $$y \in A$$ and $$\tau(y) = 0$$ if $$y \notin A$$. Since $$A$$ is not Borel, $$\tau$$ is not Borel measurable.