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...


1 Answer 1


Partial answer

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 $R$.

  • $\begingroup$ The abs.cts. part follows from DCT, right? $\endgroup$
    – MikeG
    Commented Feb 19 at 1:01
  • $\begingroup$ I do not understand DCT. Any Lipschitz function is absolutely continuous. $\endgroup$ Commented Feb 19 at 7:39
  • $\begingroup$ Sorry but why is there Lipschitzness involved in the limiting measure? $\endgroup$
    – MikeG
    Commented Feb 19 at 20:12
  • 1
    $\begingroup$ The Lipschitzness of the cumulative distribution function is a simple sufficient condition for absolute continuity. Yet, as I wrote, it is a too strong condition. $\endgroup$ Commented Feb 20 at 7:43

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