Smooth compactly supported tests are dense for finite Radon L2 measures

By [F1], given $ϵ>0$ choose $g\in C_c({\mathbb{R}}^n)$ with $\mathrm{\parallel }f-g{\mathrm{\parallel }}_{L^2(M)}<ϵ\mathrm{/}2$. Extend g by zero on ${\mathbb{R}}^n$; it is bounded and uniformly continuous. Let ${\rho }_{\delta }$ be the mollifier generated by [F3]. Since $\rho$ has compact support, for some $R<\mathrm{\infty }$ its support is contained in $B_R(0)$, and ${\sup }_x\mathrm{\mid }(g\ast {\rho }_{\delta })(x)-g(x)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }z\mathrm{\mid }\le R\delta ,x}\mathrm{\mid }g(x-z)-g(x)\mathrm{\mid }⟶0$ by uniform continuity.

The function $g\ast {\rho }_{\delta }$ is smooth by [F2], and its support lies in the sum of the compact supports of g and ${\rho }_{\delta }$, hence is compact. Since M is finite, $\mathrm{\parallel }g\ast {\rho }_{\delta }-g{\mathrm{\parallel }}_{L^2(M)}\le M({\mathbb{R}}^n{)}^{1\mathrm{/}2}\mathrm{\parallel }g\ast {\rho }_{\delta }-g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ Choose $\delta$ small enough that this is below $ϵ\mathrm{/}2$ (if $M$ is zero, every $L^2(M)$ class is zero and the conclusion is immediate). The triangle inequality gives $\mathrm{\parallel }f-g\ast {\rho }_{\delta }{\mathrm{\parallel }}_{L^2(M)}<ϵ$, proving density.