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 $X$.
Now the proof start with this phrase that I cant follow:
(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.
Can someone clarify the statement on (2)? Thank you.
P.S.: this comes from the page 76 of the book Analysis III of Amann and Escher