# Weak convergence of random measures generated by non-negative martingales?

If I have a sequence of non-negative continuous martingales $$(M_n(x))_{n\ge 1}$$ on $$x\in[0,1]$$, i.e. for each fixed $$n$$, $$M_n:[0,1]\to[0,\infty)$$ is a continuous process, and for each fixed $$x\in[0,1]$$, $$(M_n(x))_{n\ge 1}$$ is a non-negative martingale.

Then surely we have for each fixed $$x$$, a.s. we have $$M_n(x)\to M(x)$$ with some limiting stochastic process $$M$$ not necessarily continuous. But do we have the following:

Let $$\mu_n(dx)=M_n(x)dx$$ be a sequence of random measures on $$[0,1]$$, then does there need to exist some random measure $$\mu$$ on $$[0,1]$$, such that $$\mu_n$$ converges weakly to $$\mu$$ almost surely? Moreover, is there any criterion for $$\mu\ll dx$$?

Here $$dx$$ is just Lebesgue measure on $$[0,1]$$. This looks fundamental, but I didn't find it, probably due to I was not searching with right words.

I am confused how to combine the a.s. convergence of fixed $$(M_n(x))$$ to that of $$\mu_n$$ which requires far more than countable points...

For every $$a \in [0,1]$$, $$(\mu_n([0,a]))_{n \ge 0}$$ is still a non-negative martingale, hence it converges almost surely to some random variable $$L_a$$ with values in $$[0,+\infty]$$. One can set $$L_a = \liminf_{n \to +\infty} \mu_n([0,a])$$ to have $$L_a$$ defined everywhere and to have the process $$(L_a)_{a \ge 0}$$ surely non-decreasing.
By Fatou's Lemma and Fubini's theorem $$\mathbb{E}[L_1] \le \liminf_{n \to +\infty} \mathbb{E}[\mu_n([0,1])] = \int_{[0,1]} \mathbb{E}[M_n(x)] \mathrm{d}x = 1.$$ Hence $$L_1 < +\infty$$ almost surely.
Almost surely, for all $$a \in \mathbb{Q} \cap [0,1]$$, $$\mu_n([0,a]) \to L_a$$ as $$n \to +\infty$$. Since we work with non-decreasing processes, this convergence follows for all continuity point of the process $$(L_a)_{a \ge 0}$$. This provides the almost sure convergence of the measure $$\mu_n$$ as $$n \to +\infty$$, to the Stieltjes measure associated to $$(L_{a+})_{a \ge 0}$$.
A too strong condition to ensure that this limit measure is absolutely continuous is that the random variables $$\sup\{M_n(x) : x \in [0,1]\}$$ are dominated by some finite random variable $$R$$.