# Reasoning in a proof of Luzin's theorem

Luzin's theorem says:

(1) Suppose $$X$$ is a $$\sigma$$-compact metric space, $$\mu$$ is a complete Radon measure on $$X$$, and $$f\in\mathcal L_0(X,\mu, E)$$. Then for every measurable set $$A$$ with finite measure and every $$\epsilon>0$$ there is a compact $$K\subset X$$ such that $$\mu(A\setminus K)<\epsilon$$ and $$f|_K\in C(K,E)$$.

Note: above the notation $$f\in\mathcal L_0(X,\mu, E)$$ just means that $$f:X\to E$$ is $$\mu$$-measurable, where $$E$$ is a Banach space and $$X$$ is $$\sigma$$-finite. Also here $$\sigma$$-compact means that $$X$$ is locally compact and that there is a sequence of compact sets that cover $$X$$.

(2) Because $$X$$ is $$\sigma$$-compact then there is some compact $$K\subset X$$ such that $$\mu(A\setminus K)<\epsilon/2$$.
I can understand the above but not because $$X$$ is $$\sigma$$-compact, if not because $$\mu$$ is regular and then because $$\mu(A)<\infty$$ then there is some $$K\subset A$$ such that $$\mu(A)-\mu(K)=\mu(A\setminus K)<\epsilon/2$$, but I cant follow the reasoning about this property just because $$X$$ is $$\sigma$$-compact.
• @Kavi the definition on the book says that $\mu$ is a Radon measure on $X$ if $X$ is $\sigma$-compact (as explained in the question) and $\mu$ is locally finite, regular and any borel set of $X$ is $\mu$-measurable. – Masacroso Sep 25 '18 at 1:08