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

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

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.
• Could you explain what it means for a set to be $\sigma$-finite? Commented May 11, 2022 at 23:29
• @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/… Commented May 11, 2022 at 23:31
• 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? Commented May 11, 2022 at 23:41
• @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. Commented May 11, 2022 at 23:51