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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

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  • $\begingroup$ Definition of Radon measure is not very standard. What is the definition in this book? $\endgroup$ – Kavi Rama Murthy Sep 24 '18 at 23:48
  • $\begingroup$ @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. $\endgroup$ – Masacroso Sep 25 '18 at 1:08
  • $\begingroup$ You are right in saying that sigma compactness need not have been mentioned in 2) because regularity is part of the definition. $\endgroup$ – Kavi Rama Murthy Sep 25 '18 at 5:18

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