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...