(ZF + Countable Choice)

Let $\langle A,\mathcal{S} \hspace{.02 in} \rangle$ and $\langle B,\mathcal{T} \hspace{.06 in} \rangle$ be second-countable Hausdorff spaces.
Let $\Sigma$ be a sigma-algebra on $A$ such that $\mathcal{S} \subseteq \Sigma$. Let $\mu : \Sigma \to [0,+\infty]$ be an outer regular measure.
Let $X : A\to B$ be measureable. Does it follow that the pushforward measure $X_*(\mu)$ is outer regular?

If no, what if:
$\quad$ $\Sigma$ is exactly $A$'s Borel sets?
$\quad$ $\mu(A) = 1$ ?


1 Answer 1


For infinite measures $\mu$, the pushforward measure need not be outer regular, even if $\Sigma$ is the $\sigma$-algebra of Borel sets.

For a simple counterexample, let $A$ be the real line $\mathbb{R}$, and let $B$ be the rationals $\mathbb{Q}$ (with subspace topology). Let $\mu$ be Lebesgue measure. Write the rationals using an injective integer-indexed enumeration $(q_z)_{z \in \mathbb{Z}}$. Let $X$ be the function from $\mathbb{R}$ to $\mathbb{Q}$ defined by setting $X^{-1}(q_z)$ to be the half-open interval $[z,z+1)$. The inverse image of any subset of $\mathbb{Q}$ under $X$ is a countable union of half-open intervals, hence Borel. Thus the function $X$ is (Borel) measurable. But the pushforward measure $X_{\star}(\mu)$ is the counting (cardinality) measure on $\mathbb{Q}$. That is $X_{\star}(\mu)(Z) = |Z|$ for any subset $Z \subseteq \mathbb{Q}$. This is not outer regular because the measure of a nonempty open is always $\infty$.

The case of a finite measure $\mu$ has a trivially affirmative answer in the special case that the topological space $B$ is regular, because any finite measure $\nu$ on a second-countable regular space $B$ is automatically outer regular. (I'm sorry I don't know a direct reference this, though it must be standard. I have pieced it together from the following. By second-countability the restriction of $\nu$ to open sets is a continuous valuation. Any finite continuous valuation on a regular space extends, via outer measure defined using opens, to an outer-regular measure $\nu'$ on Borel sets. But $\nu$ and $\nu'$ are two finite Borel measures agreeing on open sets, and hence equal. Thus $\nu$ is outer regular. For the extension part of this argument see Theorem 4.4 of Alvarez-Manilla, "Extension of valuations on locally compact topological spaces", Topology and its Applications, 124(3):397-433 (2002).)

I do not know the answer for finite $\mu$ and non-regular $B$.


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