If probabilities $\mu_n\to \mu$ on every open interval of $I$, does $\mu_n\to \mu$ on the union of countable pairwise disjoint such intervals?

Question

I am stuck for hours with the following question which I does not succeed to prove or to disprove (not a homework problem).

Assume that $\mu$ and $\mu_n$ are Borel probabilities on $[0,1]$. Let $U = \bigcup_k I_k$, where $(I_k)_k$ is a countable family of pairwise disjoint open intervals of $[0,1]$ (open intervals of $[0,1]$ include intervals of the form $[0, \alpha)\subset I$ and $(\alpha, 1]\subset I$).

Given that $\mu_n(I) \to \mu(I)$ for every open interval of $[0,1]$ (hence in fact on every interval of $[0,1]$), prove or disprove that the hypotheses imply $\mu_n(U) \to \mu(U)$.

Note: If that can help to solve the problem, I am mainly interested in the case where the $\mu_n$ are discrete probabilities of the form $\sum_{i=1}^N \alpha_i \delta_{a_i}$, where $\delta_{a_i}$ is the dirac measure at $a_i$.

Answer 1

In the case that you only have finitely many intervals $(I_k)_k$, your desired result immediately follows from additivity of the measure and linearity of the limit: $$\mu(U) = \sum_k \mu(I_k) = \sum_k \lim_{n\to\infty} \mu_n(I_k) = \lim_{n\to\infty} \sum_k \mu_n(I_k) = \lim_{n\to\infty} \mu_n(U).$$ However, if $(I_k)_k$ is an infinite family, the interchange of limit and summation poses a problem. In this case you need more information on the $\mu_n$. One example is monotonicity, i.e. that $\mu_n(A) \leq \mu_{n+1}(A)$ for all Borel sets $A$, since in this case you can apply the Monotone Convergence Theorem (Edit: Monotonicity is actually impossible since $\mu_n$ are probability measures, as has been pointed out). Another option is to use Dominated Convergence Theorem, in which case you would need a sequence $(a_k)_k$ of positive numbers such that $\mu_n(I_k) \leq a_k$ for all $n,k$ and such that $\sum_k a_k < +\infty$. Without these assumptions, the best result I can see you obtain is that $$\mu(U) \leq \liminf_{n\to\infty} \mu_n(U)$$ by making use of Fatou's Lemma, or as was explained in another answer.

In your example where $\mu_n$ are discrete measures, write $$\mu_n = \sum_j A_{jn} \delta_{a_{jn}},$$ (assuming that your weights $A_{jn}$ and points $a_{jn}$ depend on $n$, otherwise things might simplify...) you will get $$\mu(U) = \sum_k \lim_{n\to\infty} \sum_j A_{jn} \delta_{a_{jn}}(I_k).$$ Now the limit is stuck between two sums, which means you cannot even exchange the sums. In particular, you will still get problems if your $a_{jn}$ "jump around" too much in your intervals $I_k$ for you to apply any convergence theorems.

Answer 2

oh my... I finally found the answer, which was quite simple, I can't believe.

Let $$\mu_n = {1\over n+1}\sum_{i=0}^n \delta_{i/n}.$$

Then $\mu_n((\alpha, \beta)$ tends to $\beta-\alpha$ for every $\alpha,\beta \in [0,1]$, $\alpha < \beta$. Also, $\mu_n([0,\alpha))\to \alpha$, $\mu_n((\alpha, 1])\to 1-\alpha$ and $\mu_n([0,1]) = 1$. So, denoting by $\mu$ the usual uniform Borel measure on $[0,1]$, $\mu_n$ and $\mu$ fulfil the hypotheses.

Notice also that $\mu_n(\mathbb Q\cap [0,1]) = 1$ for all $n$.

Now, cover the countable set $\mathbb Q\cap [0,1]$ by an open set $V$ of measure $\mu(V) < 1$, which is possible by a simple construction. It is well known that $V$ can be decomposed into a countable union of open intervals of $[0,1]$. But $\mu_n(V) = 1$ for all $n$, since $V$ contains $\mathbb Q \cap [0,1]$. Hence $\mu_n(V)\not\to \mu(V)$.