I'm reading about Lusin's theorem in textbook Optimal Transport for Applied Mathematicians
Let us be more precise: take a topological space $X$ endowed with a finite regular measure $\mu$ (i.e. any Borel set $A \subset X$ satisfies $\mu(A)=\sup \{\mu(K): K \subset A, K \text{ compact} \}=\inf \{\mu(B): B \supset A, B \text{ open}\}$ ). The arrival space $Y$ will be supposed to be second-countable (i.e. it admits a countable family $\left(B_i\right)_i$ of open sets such that any other open set $B \subset Y$ may be expressed as a union of $B_i$; for instance, separable metric spaces are second-countable).
Theorem (weak Lusin) - Under the above assumptions on $X, Y, \mu$, if $f: X \rightarrow Y$ is measurable, then for every $\varepsilon>0$ there exists a compact set $K \subset X$ such that $\mu(X \backslash K)<\varepsilon$ and the restriction of $f$ to $K$ is continuous.
Proof - For every $i \in \mathbb{N}$, set $A_i^{+}=f^{-1}\left(B_i\right)$ and $A_i^{-}=f^{-1}\left(B_i^C\right)$. Consider compact sets $K_i^{\pm} \subset A_i^{\pm}$such that $\mu\left(A_i^{\pm} \backslash K_i^{\pm}\right)<\varepsilon 2^{-i}$. Set $K_i=K_i^{+} \cup K_i^{-}$and $K=\bigcap_i K_i$. For each $i$ we have $\mu\left(X \backslash K_i\right)<\varepsilon 2^{1-i}$. By construction, $K$ is compact and $\mu(X \backslash K)<4 \varepsilon$. To prove that $f$ is continuous on $K$ it is sufficient to check that $f^{-1}(B) \cap K$ is relatively open in $K$ for each open set $B$, and it is enough to check this for $B=B_i$. Equivalently, it is enough to prove that $f^{-1}\left(B_i^c\right) \cap K$ is closed, and this is true since it coincides with $K_i^{-} \cap K$.
Theorem (strong Lusin) - Under the same assumptions on $X$, if $f: X \rightarrow \mathbb{R}$ is measurable, then for every $\varepsilon>0$ there exists a compact set $K \subset X$ and a continuous function $g: X \rightarrow \mathbb{R}$ such that $\mu(X \backslash K)<\varepsilon$ and $f=g$ on $K$.
Proof - First apply weak Lusin's theorem, since $\mathbb{R}$ is second countable. Then we just need to extend $f_{\mid K}$ to a continuous function $g$ on the whole $X$. This is possible since $f_{\mid K}$ is uniformly continuous (as a continuous function on a compact set) and hence has a modulus of continuity $\omega:\left|f(x)-f\left(x^{\prime}\right)\right| \leq \omega\left(d\left(x, x^{\prime}\right)\right)$ (the function $\omega$ can be taken sub additive and continuous). Then define $g(x)=\inf \left\{f\left(x^{\prime}\right)+\omega\left(d\left(x, x^{\prime}\right)\right): x^{\prime} \in K\right\}$. It can be easily checked that $g$ is continuous and coincides with $f$ on $K$.
My concerns are about the proof of the weak version. First, the author assumes that $X, Y$ are topological spaces where $Y$ is second-countable.
We have $$ \begin{align} \mu\left(X \backslash K_i\right) &= \mu\left(X \backslash (K_i^{+} \cup K_i^{-})\right) \\ &= \mu ( (X \setminus K_i^+) \cup (X \setminus K_i^-)) \\ &\le \mu (X \setminus K_i^+) + \mu (X \setminus K_i^-). \end{align} $$ Also, $\mu\left(A_i^{+} \backslash K_i^{+}\right)<\varepsilon 2^{-i}$ and $\mu\left(A_i^{-} \backslash K_i^{-}\right)<\varepsilon 2^{-i}$. How do we conclude $\mu\left(X \backslash K_i\right)<\varepsilon 2^{1-i}$?
The union of two compact sets is compact, so $K_i$ is compact. Here $X$ is not assumed to be Hausdorff. How do we conclude $K := \bigcap_i K_i$ is compact?
We have $f^{-1}\left(B_i^c\right) \cap K = K_i^{-} \cap K$. We know that $K_i^-$ is compact, but we don't know if it is closed. How do we conclude that $f^{-1}\left(B_i^c\right) \cap K$ is closed in $K$?
It seems to me $\mu$ don't need to be outer regular.
Could you please elaborate on my confusion?
