Let $X$ be a locally compact Polish space. Is the space of continuous functions with compact support dense in that of $\mu$-integrable functions

general-topologylp-spacesmeasure-theorypolish-spacesreal-analysis

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. $\|\cdot\|_{L_1}$.

Best Answer

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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