Lusin’s Theorem for Polish spaces with infinite Radon measure I’m working on the following exercise in Klenke’s Probability Theory: A Comprehensive Course (Exercise 13.1.3), which asks us to prove the following generalization of Lusin’s Theorem:

Let $\Omega$ be a Polish space, let $\mu$ be a $\sigma$-finite measure on $(\Omega, \mathcal B(\Omega))$, and let $f : \Omega \to \mathbb R$ be a map. Show that the following are equivalent:

*

*There is a Borel measurable map $g : \Omega \to \mathbb R$ with $f = g$ almost everywhere.

*For any $\epsilon > 0$, there is a compact set $K_\epsilon$ with $\mu(\Omega \setminus K_\epsilon) < \epsilon$ such that the restricted function $f|_{K_\epsilon}$ is continuous.


As stated, this exercise is wrong when $\mu(\Omega) = \infty$: if $\Omega = \mathbb R$, no compact set has a complement with finite Lebesgue measure, so it should be a closed set $K_\epsilon$.
Furthermore, $\mu$ must be more than just $\sigma$-finite. Let $\Omega = \mathbb R$, and $\mu = \sum_{q \in \mathbb Q} \delta_q$ be the counting measure of the rationals. Then $\mu$ is certainly $\sigma$-finite, but if $f$ is a Borel-measurable map and if $K \subset \mathbb R$ is closed with $\mu(K^c) < \epsilon$ for $\epsilon < 1$, then we must have $\mu(K^c) = 0$, or $K \supset \mathbb Q$. But then since $K$ is closed, $\overline{\mathbb{Q}}= \mathbb R \subset K$, so $f$ must be continuous on $\mathbb R$ in order for the claim to hold. So we need more than $\sigma$-finite.
One way to edit the exercise is to instead assume $\mu$ is Radon and modify Statement 2 like so:

*

*There is a Borel measurable map $ g : \Omega \to \mathbb R$ with $f = g$ $\mu$-a.e.

*For any subset $A \subset \Omega$ with $\mu(A) < \infty$, and for any $\epsilon > 0$, there is a compact $K_\epsilon \subset A$ such that $f|_{K_\epsilon}$ is continuous.

These statements may be shown to be equivalent, since one can show Radon measures on Polish are $\sigma$-finite (see the discussion below).
But suppose we want to show the “original” Statement 2:

For any $\epsilon > 0$, there is a closed $K_\epsilon \subset \Omega$ with $\mu(K_\epsilon^c) < \epsilon$ such that the restricted function $f|_{K_\epsilon} : K_\epsilon \to \mathbb R$ is continuous.

What conditions must we impose on the Polish space $\Omega$ with infinite Radon measure $\mu$ in order to guarantee that this is true?
 A: All references I can find have extra assumptions, but I believe they are not needed. Hopefully my argument is correct, but there is a strictly positive probability that I overlooked some tricky details. I will prove the following, which is the most general version I could manage with a Radon measure.
Theorem: Let $X$ be a topological space, $\mu$ be a Radon measure on $X$ and $f\colon X\to Y$ a function, where $Y$ is a second countable topological space. The following are equivalent:

*

*There is a Borel function $g$ such that $f=g$ almost everywhere.

*For every $\epsilon>0$ there is a closed $F\subseteq X$ such that $f_{|F}$ is continuous and $\mu(X\setminus F)<\epsilon$.

Proof: $(1)\implies $(2) Fix $\epsilon>0$ and a countable basis $\{Y_n\}_{n\in\Bbb N}$ for $Y$ and, for each $n$, find a closed $F_n$ and an open $V_n$ such that $F_n\subseteq g^{-1}(Y_n)\setminus D\subseteq V_n$ and $\mu(V_n\setminus F_n)<\epsilon/2^{n+1}$, where $D=\{x\in X\mid f(x)\neq g(x)\}$. Let $U=\bigcup(V_n\setminus F_n)$ and note that $U$ is an open set with $\mu(U)<\epsilon$. Let $F=X\setminus U$, and note that $g^{-1}(Y_n)\cap F=V_n\cap F$, hence $g_{|F}$ is continuous, but $f=g$ on $F$ so $f_{|F}$ is also continuous.
$(2)\implies(1)$ For every $n\in\Bbb N$ we can find a closed set $F_n$ such that $f_{|F_n}$ is continuous and $\mu(X\setminus F_n)<1/n$. Let $F=\bigcup F_n$, let $Z=X\setminus F$ and fix a countable basis $\{Y_n\}_{n\in\Bbb N}$ for $Y$. We have $$f^{-1}(Y_m)=(f^{-1}(Y_m)\cap Z)\cup\bigcup_{n\in\Bbb N}f^{-1}(Y_m)\cap F_n,$$
where the set on the right is Borel, since $f_{|F_n}$ is continuous, while the set on the left is contained in the Borel null set $Z$, so if we define $g:X\to Y$ to agree with $f$ on $F$ and to be constant on $Z$ we have the function we were looking for.
As an aside note that to some authors a Radon measure means a measure which is tight rather than just inner regular, meaning that for all measurable $U$ and all $\epsilon>0$ there is a compact $K\subseteq U$ with $\mu(U\setminus K)<\epsilon$. Those authors usually state Lusin's theorem with compact $K_\epsilon$ and then remark that if the measure is Borel regular instead of Radon then the theorem still holds by replacing "compact" with closed", maybe the book you're following is using this convention?
