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I'm reading this question for which I would like to clarify the theorem mentioned there. We have

  • (S1) Let $X$ be a locally compact Hausdorff space. Then the space of continuous functions with compact support is dense in that of continuous functions vanishing at infinity w.r.t. $\| \cdot \|_\infty$. ref.

  • (S2) Let $X$ be $\sigma$-compact, locally compact Hausdorff space and $\mu$ is a Radon measure on $X$. Then the space of continuous functions with compact support is dense in that of $\mu$-integrable functions w.r.t. $\|\cdot\|_{L_1}$. ref

Does (S2) still hold if we drop the $\sigma$-compactness condition? If not, does below statement hold?

Let $X$ be a locally compact Polish space and $\mu$ is a Borel measure on $X$. Then the space of continuous functions with compact support is dense in that of $\mu$-integrable functions w.r.t. $\|\cdot\|_{L_1}$.

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Yes, S2 holds without the assumption that $X$ be $\sigma$-compact. To see this, observe that the space of simple functions is dense in $L^1(\mu)$, so it suffices to show that simple functions can be approximated by functions in $C_c(X)$. Moreover, if $g \in L^1$, the set $\{x \in X \, : \, g(x) \neq 0 \}$ is $\sigma$-finite, so in fact it suffices to show that a simple function $\chi_A$ with $\mu (A) < \infty$ can be approximated by an element of $C_c(X)$.

Let $\epsilon > 0$. Since $\mu$ is Radon, for a $\mu$-finite set $A$ there exists a compact set $K$ and an open set $U$ such that $K \subset A \subset U$ and $\mu (U \setminus K) < \epsilon$. By Urysohn's lemma, one can choose an $f \in C_c(X)$ with $\chi_K \leq f \leq \chi_U$. It follows that $$\|f - \chi_A \|_{L^1} \leq \mu (U \setminus K) < \epsilon$$ as desired.

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  • $\begingroup$ Could you explain what it means for a set to be $\sigma$-finite? $\endgroup$
    – Akira
    Commented May 11, 2022 at 23:29
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    $\begingroup$ @Akira A set $A$ is $\sigma$-finite for $\mu$ if it admits a decomposition $A = \bigcup_{i=1}^\infty E_i$ such that $\mu (E_i) < \infty$ for each $i$. See e.g. math.stackexchange.com/questions/2821538/… $\endgroup$ Commented May 11, 2022 at 23:31
  • $\begingroup$ Thank you so much! $\endgroup$
    – Akira
    Commented May 11, 2022 at 23:33
  • $\begingroup$ A Borel measure $\mu$ on a Hausdorff topological space $X$ is called Radon if $\mu$ is tight and locally finite. In (S2), $X$ is locally compact, so $\mu$ is also outer regular. Could you confirm if my understanding is correct? $\endgroup$
    – Akira
    Commented May 11, 2022 at 23:41
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    $\begingroup$ @Akira In fact, more is true. Since you have a locally compact Polish space, every open set is $\sigma$-compact, so that any Borel measure that is locally finite is (inner and outer) regular. See Theorem 7.8 in Folland's Real Analysis. $\endgroup$ Commented May 11, 2022 at 23:51

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