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.

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jul 10 at 13:06
  • $\begingroup$ @TobiasFritz $\tau$ may not even be well-defined in that case, since there’s no reason $\{x \in X: \pi(x) = y\}$ is measurable. $\endgroup$
    – David Gao
    Commented Jul 10 at 17:12

1 Answer 1


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.


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