If $\lim_n \mu_n ([0, r]) = \mu ([0, r])$ for all $r \in D$, then $\mu_n \to \mu$ in distribution

Question

Let $T>0$ and $\mu, \mu_n$ be Borel probability measures on the closed interval $[0, T]$. Let $D$ be a dense subset of $[0, T]$. I would like to prove that

Theorem If $\lim_n \mu_n ([0, r]) = \mu ([0, r])$ for all $r \in D$, then $\mu_n \to \mu$ in distribution.

Could you elaborate on how to finish my below failed attempt?

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Proof Fix $r \in [0, T]$ such that the c.d.f. of $\mu$ is continuous at $r$. We need to prove that $\lim_n \mu_n ([0, r]) = \mu ([0, r])$. We fix $\varepsilon>0$. We need to show that there is $N>0$ such that here is $r' \in [0, T]$ such that $$ |\mu_n ([0, r]) - \mu ([0, r])| < \varepsilon \quad \forall n>N. $$

There is $r' \in D$ such that $$ |\mu ([0, r]) - \mu ([0, r'])| < \varepsilon/2. $$

It suffices to prove that there is $M>0$ such that $$ |\mu_n ([0, r]) - \mu ([0, r'])| < \varepsilon/2 \quad \forall n>M. $$

Answer 1

Shortcut:

Let $x\in\left(0,T\right)$ such that $\mu\left(\left\{ x\right\} \right)=0$.

Then for any fixed $\epsilon>0$ we then can find $r_{\epsilon},s_{\epsilon}\in D$ with $r_{\epsilon}<x<s_{\epsilon}$ and $\mu\left(\left(r_{\epsilon},s_{\epsilon}\right]\right)<\epsilon$.

This with: $$\mu_{n}\left(\left[0,r_{\epsilon}\right]\right)\leq\mu_{n}\left(\left[0,x\right]\right)\leq\mu_{n}\left(\left[0,s_{\epsilon}\right]\right)$$ for every $n$ and consequently by taking the limit:

$$\mu\left(\left[0,r_{\epsilon}\right]\right)\leq\liminf\mu_{n}\left(\left[0,x\right]\right)\leq\limsup\mu_{n}\left(\left[0,x\right]\right)\leq\mu\left(\left[0,s_{\epsilon}\right]\right)$$

So: $$\limsup\mu_{n}\left(\left[0,x\right]\right)-\liminf\mu_{n}\left(\left[0,x\right]\right)<\epsilon$$ This for every $\epsilon>0$ hence making clear that $\mu_{n}\left(\left[0,x\right]\right)$ converges.

The inequalities: $$\mu\left(\left[0,r_{\epsilon}\right]\right)\leq\mu\left(\left[0,x\right]\right)\leq\mu\left(\left[0,s_{\epsilon}\right]\right)$$make clear that $\mu\left(\left[0,x\right]\right)$ is the only candidate for being the limit.

Answer 2

Fix $r \in [0, T]$ such that the c.d.f. of $\mu$ is continuous at $r$. We need to prove that $\lim_n \mu_n ([0, r]) = \mu ([0, r])$. There is $r_n > r_{n+1} > r$ such that $r_n \in D$ for all $n \in \mathbb N$ and $r_n \downarrow r$ and $$ 0 \le \mu_n ([0, r_n]) - \mu_n ([0, r]) <\frac{1}{n}. $$

It follows that $$ \lim_n \big ( \mu_n ([0, r_n]) - \mu_n ([0, r]) \big ) = 0. $$

It suffices to prove that $$ \lim_n \mu_n ([0, r_n]) = \mu ([0, r]). $$

1. First, $$ \mu_n ([0, r_n]) \le \mu_n ([0, r_m]) \quad \forall m \le n. $$

So $$ \lim_n \mu_n ([0, r_n]) \le \lim_n \mu_n ([0, r_m]) =\mu ([0, r_m]) \quad \forall m \in \mathbb N. $$

Hence $$ \lim_n \mu_n ([0, r_n]) \le \lim_m \mu ([0, r_m]) = \mu ([0, r]). $$

2. Second, fix $\varepsilon>0$. There is $r_\varepsilon \in D$ such that $r_\varepsilon < r$ and $$ 0 \le \mu ([0, r]) - \mu ([0, r_\varepsilon]) <\varepsilon. $$

Such $r_\varepsilon$ is exists because the c.d.f. of $\mu$ is continuous at $r$. Then $$ \lim_n \mu_n ([0, r_n]) \ge \lim_n \mu_n ([0, r_\varepsilon]) = \mu ([0, r_\varepsilon]) > \mu ([0, r]) - \varepsilon. $$

Take the limit $\varepsilon \downarrow 0^+$. Then $$ \lim_n \mu_n ([0, r_n]) \ge \mu ([0, r]). $$

This completes the proof.