I'm struggling with producing a proof of the following result:

Let $X = \overline{\mathbb{C}}$ be the Riemann sphere, and consider $M(X)$ the space of finite Borel measures on $X$ with norm given by the total variation. Let $\mathcal{H}$ be a basis of open sets for the standard euclidean topology on $\overline{\mathbb{C}}$, such that for any $H \in \mathcal{H}$, $\mu(\overline{H} \setminus H) = \mu(\overline{H} \setminus \text{int}(H)) = 0$ where $\mu \in M(X)$ is some fixed positive measure. Then for any sequence of positive measures $\mu_{n} \in M(X)$ ,, $\mu_{n}(H) \rightarrow \mu(H)$ for all $H \in \mathcal{H}$ implies $\mu_n$ converges weakly to $\mu$.

Supposedly one can proceed by 'uniformly approximating $f \in C(X)$' by linear combinations of indicator functions on the members of $\mathcal{H}$, but I do not see how to accomplish this.

Thank you in advance.


2 Answers 2


I've started writing this before you posted your answer, so I'll post this anyway since it took me some time to write in the hopes it might still benefit someone else.

First, note that $\nu(H)=\int\mathbf{1}_Hd\nu$ for every measure. I'm assuming you want to show that $\int_X f d\mu_n\to \int_Xfd\mu$ for every $f\in C(X)$.

Notice that if you can find for every $\epsilon>0$ an $f_\epsilon= \sum_{\ell=1}^{N_\epsilon} c_\ell \cdot \mathbf{1}_{G_\ell}$ such that $\Vert f-f_\epsilon\Vert_\infty <\epsilon$, then you can write

$$\int_X fd\mu-\int_X fd\mu_n= \Big(\int_X fd\mu- \int_X f_\epsilon d\mu \Big)+ \Big(\int_X f_\epsilon d\mu- \int_X f_\epsilon d\mu_n \Big)+ \Big(\int_X f_\epsilon d\mu_n- \int_X f d\mu_n \Big)\leq $$

And we can then deduce that

$$\Big\vert \int_X fd\mu-\int_X fd\mu_n \Big\vert \leq \epsilon(\mu(X)+\mu_n(X))+ \Bigg\vert \int_X f_\epsilon d\mu-\int_X f_\epsilon d\mu_n \Bigg\vert.$$

We can also deduce that

$$ \Bigg\vert \int_X f_\epsilon d\mu-\int_X f_\epsilon d\mu_n \Bigg\vert \leq \sum_{\ell=1}^{N_\epsilon} \vert a_\ell \vert \cdot \vert\mu(G_\ell)-\mu_n(G_\ell) \vert. $$

We can choose $G_\ell$ from the algebra generated by $\mathcal{H}$ since $X$ is compact and $f$ is uniformly continuous. By the assumption on $\mu$, we can conclude that $\mu_n(G)\to \mu(G)$ for every $G$ in the set algebra generated by $\mathcal{H}$. This last thing does require some work which I skip.

Finally one has to apply the measure convegrence assumption on finitely many sets in $\mathcal{H}$, including $X$ itself, and obtain that

$$ \Big\vert \int_X fd\mu-\int_X fd\mu_n \Big\vert < 4\epsilon. $$


So I think given compactness, the above should imply weak convergence for any basis of open sets at all, regardless of whatever condition we impose on it. My argument is as follows:

First note that by compactness, we have that $\mu_{n}(X),\mu(X)$ are uniformly bounded, say by $C > 0$. Set $\mu_{0} = \mu$. This means that if $\phi_{m} \rightarrow g$ uniformly on $X$, $| \int \phi_m d \mu_{n} - \int g d \mu_{n} | \leq C ||\phi_{m} - g||_{u} \rightarrow_{m} 0$, independent of $n = 0,1,...$

Now WLOG suppose for $f \in C(X)$, we have $0 \leq f \leq 1$, define $\phi_{m} = \frac{1}{2^m} \sum_{j=1}^{2^m -1} \mathbf{1}_{f^{-1}(\frac{j}{2^m},+\infty)} = \sum_{p \geq 1} c^{m}_{p} \mathbf{1}_{U^{m}_{p}}$ for some $c^{m}_{p} > 0$ and $U^{m}_{p} \in \mathcal{H}$.

Then $\int \phi_{m} d \mu_{n} \rightarrow_{n \rightarrow \infty} \int \phi_{m} d \mu$ for all $m$ by hypothesis, however the convergence depends on $m$. Moreover since $f$ is bounded $0 \leq \phi_{m} \uparrow f$ uniformly on $X$.

Let $M$ be such that for $m \geq M$, and some fixed $\epsilon > 0$, we have $|\int \phi_{m} d\mu_{n} - \int f d\mu_{n} | < \frac{\epsilon}{3}$ for all $n$.

Choose $N_0$ such that for $n \geq N_0$, one has $|\int \phi_{M} d\mu_{n} - \int \phi_{M} d \mu | < \frac{\epsilon}{3}$.

Then $|\int f d\mu_{n} - \int f d\mu| \leq |\int f d\mu_{n} - \int \phi_{M} d\mu_{n}| + |\int \phi_{M} d\mu_{n} - \int \phi_{M} d\mu| + |\int \phi_{M} d\mu - \int f d \mu| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} < \epsilon$ whenever $n \geq N_0$, so we have proved $\lim_{n \rightarrow \infty} \int f d\mu_{n} = \int f d\mu$ as desired.

  • $\begingroup$ I think the boundedness of $\mu_n(X)$ comes from the measure convergence assumptions. You can have infinite Borel measures on compact spaces, and you're not assuming that $\mu_n$ are regular. $\endgroup$ May 26 at 15:01
  • $\begingroup$ I wrote $M(X)$ is the space of finite borel measures on $X = \overline{\mathbb{C}}$, which automatically makes $\mu$ regular since $X$ is $\sigma$-compact, basically by Riesz, also I think weak $^{\ast}$ convergence is always talked about with respect to measures on a locally compact hausdorff space, in which case $M(X) = C_{0}(X)^{\ast}$ are all finite measures, because their total variations are the same as the norms of the corresponding bounded linear functionals. So in such a situation $\mu(X) $ finite seems totally safe to assume. $\endgroup$ May 26 at 15:12
  • $\begingroup$ You mean the Riesz-Markov Kakutani theorem? It talks about Radon measures not Borel measures. The counting measure would be Borel and not finite. $\endgroup$ May 26 at 15:16
  • $\begingroup$ Anyway, to flesh out what I meant by 'Compact -> $\mu_n(X)$ uniformly bounded', the argument I had in mind was : Let $H_1,..,H_K \in \mathcal{H}$ cover $X$. Then $\mu(X) \leq \sum_{j=1}^k \mu(H_j) = M$, and $\mu_{n}(X) \leq \sum_{j=1}^k \mu_{n}(H_j) \leq M + 1$ for $n$ sufficiently large, say for $n \geq N_0$. But then $\mu_1(X),...,\mu_{N_0 -1}(X)$ is bounded above by say $N$, so $\mu_n(X) \leq \text{max}(M+1,N)$ for all $n$ $\endgroup$ May 26 at 15:16
  • $\begingroup$ The problem with your argument, is that you can take one finite measure $\mu$ and $\mu_n=n\cdot \mu$ would all be bounded, but not uniformly. $\endgroup$ May 26 at 15:19

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