Given a set $X$, an outer measure on $X$ is a function $\mu:\wp(X)\to [0,+\infty]$ such that $\mu(\varnothing)=0$ and $\mu(A)\leq \sum_{j=1}^\infty \mu(A_j)$, whenever $A\subset \bigcup_{j=1}^\infty A_j$. A set $A\subset X$ is said to be measurable if $\mu(S)=\mu(S\cap A)+\mu(S-A)$, for all $S\subset X$. The collection of $\mu$-measurable sets is a $\sigma$-algebra on $X$, so the restriction of an outer measure $\mu$ on the $\mu$-measurable subsets of $X$ is a measure in the usual sense.

If $(X,d)$ is a metric space, we say that an outer measure $\mu$ on $X$ is a Radon measure, if

  1. $\mu$ is Borel-regular (i.e. every Borel set in $X$ is measurable and, for every $A\subset X$, there is some Borel set $B\supset A$ such that $\mu(A)=\mu(B)$);
  2. $\mu(K)<\infty$, for every $K\subset X$ compact;
  3. $\mu(A)=\inf\{\mu(U)\,:\, A\subset U, U \text{ open}\}$ for every $A\subset X$;
  4. $\mu(U)=\sup\{\mu(K)\,:\, K\subset U, K \text{ compact}\}$ for every open $U\subset X$;

Denote the collection of all Radon measures on $(X,d)$ by $\mathcal{R}(X)$. We say that a sequence $\mu_n$ in $\mathcal{R}(X)$ weakly converges to $\mu\in \mathcal{R}(X)$ if $$\lim_{n\to \infty}\int_Xf\,d\mu_n=\int_Xf\,d\mu,$$ for every $f\in \mathcal{C}_c(X)$ (the set of the continuous functions $f:X\to \mathbb{R}$ with compact support).

I want, if possible, to endow $\mathcal{R}(X)$ with a topology such that the notion of convergence induced by it coincides with the weak convergence defined above. Following the idea given here, I've tried the following:

For each $f\in \mathcal{C}_c(X)$, $x\in \mathbb{R}$ and $\delta>0$, define $$ U_{f,x,\delta}=\left\{\mu\in \mathcal{R}(X)\,:\,\left|\int_Xf\,d\mu-x\right|<\delta \right\},$$ and consider $\scr B$ the collection of all such $U_{f,x,\delta}$'s. Now try to show that

  1. $\scr B$ is a basis for a topology on $\mathcal{R}(X)$;
  2. Such topology has the required property (its convergence is given by the weak convergence).

To show 1., suppose $\mu\in U_{f,x,\delta}\cap U_{g,y,\epsilon}$. Then we need to find $h\in \mathcal{C}_c(X)$, $z\in \mathbb{R}$ and $\varepsilon>0$ such that $$ \mu\in U_{h,z,\varepsilon}\subset U_{f,x,\delta}\cap U_{g,y,\epsilon}.$$

Following the definitions, I've got that $$\left|\int_X(f-g)\,d\mu-(x-y)\right|<\delta+\epsilon,$$ therefore $\mu\in U_{f-g,x-y,\delta+\epsilon}$, but, I was not able to show that $\nu\in U_{f-g,x-y,\delta+\epsilon}\implies \nu\in U_{f,x,\delta}\cap U_{g,y,\epsilon}$ (I think this choice of $h$, $z$ and $\varepsilon$ is probably wrong).

And I really don't know what else to do. The Wikipedia link above defines such topology for the collection of probability measures (those with $\mu(X)=1$), not Radon ones... Such a construction fails for Radon measures?


I only know some basic functional analysis, and I will try to answer question two.

Instead of $N_{f,x,\epsilon}$ let us use $$N_{f,\epsilon}(\nu):= \bigg\{\mu\in \mathcal R(X) : \left| \int_X fd\mu - \int_X f d\nu \right|< \epsilon \bigg\}.$$ $N_{f,\epsilon}(\nu)$ defines an open neighborhood of the element $\nu\in \mathcal{R}(X)$.

Now I want to say that $\{N_{f,\epsilon}(\nu)\}$ defines a local subbasis for the neighborhood system at $\nu$, i.e. $$\mathcal B(\nu) := \{\text { finite intersections of elements from $\{N_{f,\epsilon}(\nu)\}$}\}$$ will form a local basis at $\nu$.

We check that this topology will give us the weak convergence define by limits.

  1. Suppose $\lim_n \int_X f d\nu_n = \int_X f d\nu$ for each $f$, and let $N_{f_1, \epsilon_1}\cap \cdots \cap N_{f_k, \epsilon_k}\in \mathcal B(\nu)$. By the definition of limits, there exist $L_i$'s such that for all $n\geq L_i$, we have $$\left| \int_X f_i d\nu_n - \int_X f_i d\nu\right| <\epsilon_i.$$ Define $L = \max_{i=1, \cdots, k}\{L_i\}$, then for all $n\geq L$, we have $\nu_n \in N_{f_1, \epsilon_1}\cap \cdots \cap N_{f_k, \epsilon_k}$.

  2. Conversely, given a sequence $\nu_n$ and for each $N_{f,\epsilon}(\nu)\in \mathcal B(\nu)$, there exists an $L$ such that for $n\geq L$, we have $$\left| \int_X f d\nu_n - \int_X f d\nu\right| <\epsilon,$$ this is the exact definition of $\lim_n \int_X f d\nu_n = \int_X f d\nu$.

So now question one is asking, when will $\{N_{f,\epsilon}(\nu)\}$ form a local basis instead of a sub subbasis. So given $N_{f_1, \epsilon_1}(\nu)\cap N_{f_2, \epsilon_2}(\nu)$ does there exist a function $g$ and a $\delta$ such that $$N_{g,\delta}(\nu) \subset N_{f_1, \epsilon_1}(\nu)\cap N_{f_2, \epsilon_2}(\nu).$$

  • $\begingroup$ Nice idea of working with subbasis. Maybe I don't need to worry about question 1. $\endgroup$ – Ders Mar 25 '18 at 18:30

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