Second Countability in Lusin's Theorem

Question

A version of Lusin's theorem states that

If $\mu$ is a finite Radon measure on $X$, and $Y$ is a second countable topological spaces, then for any Borel-measurable function $f:X\to Y$ and any $\epsilon>0$, there exists a compact set $K\subseteq X$ such that $\mu(X\setminus K)<\epsilon$ and $f|_K$ is continuous.

I want to know if there is a counterexample when $Y$ is not second countable.

Answer 1

Let $X\triangleq([0,1],\tau_1)$ and $Y\triangleq([0,1],\tau_2)$ where $\tau_1$ is the usual topology, and $\tau_2$ is induced from the Sorgenfrey topology (that is $\tau_2$ is generated by sets of the form $[a,b)\cap[0,1]$). Note that $Y$ is not second countable. Consider the Lebesgue measure $\mu\triangleq\mathcal{L}^1|_{[0,1]}$ on $X$.

Now consider the identity map $i:X\to Y$.

One can show that $\mathcal{B}([0,1],\tau_2)=\mathcal{B}([0,1],\tau_1)$, and so $i$ is Borel-measurable.

To show that Lusin's theorem does not hold it suffices to note that $Y$ is totally disconnected, whereas $X$ is connected. Hence, $i$ is cannot be continuous when restricted to any nontrivial open subset of $X$.