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?

  • $\begingroup$ 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. $\endgroup$ Commented Oct 27, 2019 at 16:06
  • $\begingroup$ Probably it's just a silly question. I have to think better about it. $\endgroup$
    – Bob
    Commented Oct 27, 2019 at 16:32

1 Answer 1


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.


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