3

Let $\Theta$ be a metric space, and let $\{\mu_\theta\}_{\theta \in \Theta}$ be a collection of probability measures over the Borel $\sigma$-algebra $\mathcal B(X)$ of some metric space $(X, \rho)$. Further, suppose that for every $\theta_0 \in \Theta$, $$\lim_{\theta \to \theta_0}\sup_{A \in \tau} |\mu_{\theta_0}(A) - \mu_\theta(A)| = 0$$ where $\tau$ denotes the collection of open sets in $X$. Does it follow that the total variation $$\lim_{\theta \to \theta_0}\sup_{A \in \mathcal B(X)} |\mu_{\theta_0}(A) - \mu_\theta(A)| = 0$$ also converges to zero?

Some information that I think may be helpful:

I believe this question can be seen as a generalization of If two Borel measures coincide on all open sets, are they equal?

Rather than coinciding, for any $\theta$ in some sufficiently small neighborhood of $\theta_0$, we have that the measures $\mu_\theta$ almost coincide with $\mu_{\theta_0}$ on all open sets. We need to show that we can find a similar neighborhood where the measures almost coincide on all measurable subsets.

In the above question, it was resolved that if $X$ is a union of an increasing sequence of open sets on which the two measures are finite (which is the case for us), then the answer is "yes" due to the monotone class theorem. I'm not familiar with this theorem, but perhaps a generalization of that argument will work in this case.

aduh
  • 8,894
ndurvasula
  • 149

1 Answers1

2

An even stronger result holds.

Let $A \in \mathcal B(X)$ and $\theta \in \Theta$. Because any finite Borel measure on a metrizable topological space is outer regular, for each $n \in \mathbb N$ there is an open $U_n \supset A$ such that $\mu_{\theta_0}(A \triangle U_n)<1/n$ and an open $V_n \supset A$ such that $\mu_\theta(A \triangle V_n)<1/n$. Let $W_n = U_n \cap V_n \in \tau$. Then,

\begin{align} |\mu_{\theta_0}(A) - \mu_\theta(A)| &\leq \mu_{\theta_0}(A \triangle W_n) + \sup_{W \in \tau}|\mu_{\theta_0}(W) - \mu_\theta(W)| + \mu_\theta(A \triangle W_n)\\ &\to_{n \to \infty} \sup_{W \in \tau}|\mu_{\theta_0}(W) - \mu_\theta(W)|. \end{align}

It follows from this that $$\sup_{A \in \mathcal B(X)}|\mu_{\theta_0}(A) - \mu_\theta(A)| \leq \sup_{A \in \tau}|\mu_{\theta_0}(A) - \mu_\theta(A)|,$$ and the reverse inequality is trivial, so $$\sup_{A \in \mathcal B(X)}|\mu_{\theta_0}(A) - \mu_\theta(A)| = \sup_{A \in \tau}|\mu_{\theta_0}(A) - \mu_\theta(A)|.$$

aduh
  • 8,894
  • Thanks! A (hopefully) quick follow-up: if $\tau$ instead consisted of the collection of open balls (as opposed to the open sets), does this results still hold, perhaps given some assumptions on the space $X$? – ndurvasula Jan 27 '21 at 17:14
  • Sorry -- final followup -- if $X$ is some subset of $\mathbb R^n$ and $\tau$ consisted of the collection of Jordan-measurable subsets (subsets that have boundary with zero Lebesgue measure), would the statement (or the weaker limit statement) still hold? – ndurvasula Jan 27 '21 at 19:50