# Are metric continuous measures setwise sequentially dense in finite Borel measures?

Let $$(\mathcal{X},d)$$ be a complete separable metric space. Say that a Borel measure $$\sigma$$ of $$(\mathcal{X},d)$$ is a metric continuous measure if for each $$x\in\mathcal{X}$$ the function $$(0,+\infty)\to [0,+\infty], r\mapsto\sigma(B_r(x))$$ is continuous, where $$B_r(x)$$ is the $$d$$ open ball centered in $$x$$ of radius $$r$$.

If $$\mu$$ is a finite Borel measure of $$(\mathcal{X},d)$$, is it true that there exists a sequence $$(\mu_n)_{n\in\mathbb{N}}$$ of finite continuous measure that converges setwise to $$\mu$$, i.e. such that for each Borel set $$E$$ of $$(\mathcal{X},d)$$ we have that $$\mu_n(E)\to\mu(E), n\to+\infty?$$

I initially tried with the simpler case where $$(\mathcal{X},d)$$ is some Euclidean space $$\mathbb{R}^m$$, trying to use as an approximating sequence a mollified sequence of $$\mu$$, i.e. $$\mu_n := \eta_{1/n}*\mu$$ where $$(\eta_\varepsilon)_{\varepsilon>0}$$ is the canonical mollifier of $$\mathbb{R}^m$$. However, using this strategy I find clear only that for each bounded continuous function $$f:\mathbb{R}^m\to\mathbb{R}$$ we have that $$\int_{\mathbb{R}^m}f\operatorname{d}\mu_n\to\int_{\mathbb{R}^m}f\operatorname{d}\mu, n\to+\infty$$, while what we are looking for is the same result for $$f=\chi_E$$ where $$E$$ is an arbitrary Borel set of $$(\mathcal{X},d)$$.

Any help?

• What if $(\mathcal{X}, d)$ is the discrete metric space on a countable set $\mathcal{X}$? Then the only continuous measure is the zero measure, and so, no non-trivial measure can be approximated by continuous measures. – Sangchul Lee Oct 27 '19 at 16:06
• Probably it's just a silly question. I have to think better about it. – Bob Oct 27 '19 at 16:32

If your metric space contains an isolated point $$p$$ and at least one other point, a unit mass at $$p$$ can't be approximated by continuous measures.