# weak convergence + bounded second moment implies convergence of the moment?

Let $$\{\mu_N\}$$ be a sequence of random measures which converges almost surely in the weak sense to a deterministic measure $$\mu$$ with impact support.

The weak convergence does not necessarily imply the convergence of the moments. But, my question is that, if we add the assumption that the second moment of $$\mu_N$$ is almost surely bounded, can we deduce that the second moment of $$\{\mu_N\}$$ converges to the second moment of $$\mu$$?

Suppose $$\mu_N$$ assigns mass $$1/(2N^2)$$ to each of $$\pm N$$ and mass $$1-1/N^2$$ to $$0$$. Then $$\mu_N$$ (which have second moment equal to $$1$$) converge weakly to the dirac measure at $$0$$.

• What if we add the assumption that the support is almost surely bounded? Do we have the result or not? Jun 3 at 15:17
• Yes, since in that case $x^2$ is a bounded continuous function on the support. Jun 3 at 15:49