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Let $(E,d)$ be a metric space and $\Gamma\subseteq C_b(E)$ be bounded, i.e. $$c:=\sup_{f\in\Gamma}\left\|f\right\|_\infty<\infty,$$ and equicontinuous, i.e. $$\forall x\in E:\forall\varepsilon>0:\exists\text{neighborhood }N\text{ of }x:f(N)\subseteq B_\varepsilon(f(x)).$$

Let $(\mu_t)_{t\in I}$ be a net of probability measures on $\mathcal B(E)$ weakly converging to a probability measure $\mu$ on $\mathcal B(E)$.

Let $\varepsilon>0$. By assumption, for each $x\in E$, we can find an open neighborhood $N_x$ of $x$ such that $f(N)\subseteq B_\varepsilon(x)$, but why can we choose $N_x$ such that $\mu(N_x)=0$?

The claim is made in the proof of Theorem 2.2.8 in Bogachev's Weak Convergence of Measures:

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BTW: Can anyone explain to me how we obtain the claim for Radon measures on general metric spaces from the result of arbitrary measures on separable metric spaces? The definition of Radon is as follows:

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0xbadf00d
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  • $\mu(\partial{N}_x)=0$, with uncountably many choices for the diameter of $N_x$, only countably many of these can have a boundary with positive measure. – Edwin Franks Dec 15 '20 at 09:43
  • @EdwinFranks Can you explain why this holds in more detail? – 0xbadf00d Dec 15 '20 at 10:09
  • For a radon probability measure $\mu$ there exists a compact set $K$ with $\mu(K)>1-\epsilon$ for all $\epsilon>0$. Working with $K$ overcomes the problem of an open cover with only an uncountable sub cover. – Edwin Franks Dec 15 '20 at 10:11
  • @EdwinFranks Your second comment is related to the last paragraph of my question (and not related to your first comment), or am I missing something? – 0xbadf00d Dec 15 '20 at 10:16
  • @Oxbadf00d The open neighborhoods $N_x$ for which $|f(y)-f(x)|<\epsilon$ for all $f\in\Gamma$ can be taken as a ball of radius $\delta_x>0$. If $\mu(\partial N_x)>0$ than a smaller ball will work. The boundaries of the smaller balls are disjoint and there are uncountably many. If each of these had positive measure then the measure of the original ball would be infinite. Any sum of uncountably many positive numbers is infinite. – Edwin Franks Dec 15 '20 at 10:22
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    @EdwinFranks (a) I'm not sure if I really understood why we can assume that $\mu(\partial N_x)=0$. Please take a look at the question I've asked for this: https://math.stackexchange.com/q/3956534/47771. Do you agree with my proof? (b) I'll think about the separability question tomorrow, but at first glance: Isn't it sufficient to assume that the measures are tight (instead of even Radon)? The crucial point should be that the variation of the measures are $<\varepsilon$ outside a compact subset and that each compact space is totally bounded. Or am I missing something? – 0xbadf00d Dec 20 '20 at 20:28
  • @Oxbadf00d One can assume that $\mu(\partial N_x)=0$ because there uncountably many choices for $N_x$ which have pairwise disjoint boundaries. If each of those boundaries had positive measure then the measure of their union would be infinite, as it would be the sum of the measures of the individual boundaries. If uncountably many had positive measure than for some $n$ the number having measure less than or equal to $1/n$ but greater than $1/(n+1)$ would have to be ucountably infinite. Otherwise the set of boundaries with positive measure would only be countable. – Edwin Franks Dec 21 '20 at 00:46
  • @Oxbadf00d A subset of measures $\mathcal{F}$ is tight means that the measures $\mu\in\mathcal{F}$ are uniformly radon, so all the measures in $\mathcal{F}$ are radon. – Edwin Franks Dec 21 '20 at 00:51
  • @EdwinFranks (a) So, do you agree to my proof in the other question? (b) No, it's the other way around: If $\mathcal F$ is (uniformly) Radon, it is (uniformly) tight. See the definitions I gave in this question. – 0xbadf00d Dec 21 '20 at 05:03
  • @EdwinFranks I was able to show the desired claim in a slightly more general scenario: It holds whenever $(\mu_t){t\in I}$ is bounded in total variation norm and (uniformly) tight and $\mu$ is tight. The metric space does then not need to be separable. However, I need to assume that the $\mu_t$ and $\mu$ are nonnegative. Actually, the only place where I need this assumption is to conclude $\lim{t\in I}|\mu_t-\mu|(B)\to0$ for all measurable $B$ with $\mu(\partial B)=0$. Does this hold for arbitrary finite signed measures as well? https://math.stackexchange.com/q/3956966/47771. – 0xbadf00d Dec 21 '20 at 08:15
  • @Oxbadf00d I am a bit onfused, I thought from the definitions given that tight was a property of the set $\mathcal{F}$ and not a property of an individual measure. As far as signed measures, I think that weak convergence means that the net ${f\mu_\alpha}$ converges weakly to $f\mu$ for suitable functions $f$. If we let $h=\mathrm{d}|\mu|/\mathrm{d}\mu$ be the Radon-Nikodym derivative of $|\mu|$ then I think (2.2.6) will hold with $\mu_\alpha$ and $\mu$ replaced by $h\mu_\alpha$ and $|\mu|$. Which should imply (2.2.6), as I think $|h|_\infty=1$. – Edwin Franks Dec 21 '20 at 11:49
  • @EdwinFranks I say that an individual measure $\mu$ is tight if the family ${\mu}$ is tight. – 0xbadf00d Dec 22 '20 at 09:20

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