$\newcommand{\mr}{\mathscr}$
Definitions.
Let $X$ be a compact metric space and $\mr P(X)$ be the set of all the probability measures on $X$. We say that a sequence $(\nu_n)$ in $\mr P(X)$ converges to $\nu\in \mr P(X)$ in the weak* sense if for all continuous functions $f:X\to \mathbb R$, we have that $$\int_Xf \ d\nu_n \to \int_X f\ d\nu$$
We say that a sequence $(\nu_n)$ in $\mr P(X)$ converges to $\nu$ strongly if for all Borel sets $A$ in $X$ we have $$\nu_n(A)\to \nu(A)$$
Question.
Suppose we are given a sequence $\mu_n$ in $\mr P(X)$. Define the averages $$\nu_n = \frac{\mu_1 +\cdots +\mu_n}{n}$$ for each $n\geq 1$. Assume that the sequence $\nu_n$ converges to a probability measure $\nu$ in the weak* sense.
Question Does it follow that $\nu_n$ actually converges to $\nu$ strongly?
In other words, do we get strong convergence from weak convergence in this special situation?
(Please Ignore: It is clear that for any Borel set $E$ we have $(\nu_n(E))$ is a Cauchy sequence. So it converges to something.)
(Thanks to @Davide Giraudo for pointing that the above was a hasty remark.)
The answer is no: let $(m_i)$ be a sequence of probability measure which converges weakly to $\nu$ but not strongly. Let $(N_k)_{k\geqslant 1}$ be a sequence of integers such that $$ N_1=1, N_{j+1}=2jN_j . $$ For $n$ such that $N_k+1\leqslant n\leqslant N_{k+1}$, we define $\mu_n:=m_k$. In this way, $\mu_n\to \nu$ weakly. We have $$\nu_{N_{j+1}}=\frac 1{N_{j+1}}\sum_{k=1}^{j}\sum_{n=N_k+1}^{N_{k+1}}\mu_n =\frac 1{N_{j+1}}\sum_{k=1}^{j}\left(N_{k+1}-N_k\right)m_k =m_j- \frac{N_j}{N_{j+1}} m_j+\frac 1{N_{j+1}}\sum_{k=1}^{j-1}\left(N_{k+1}-N_k\right)m_k. $$ For any Borel subset $E$ of $X$, $$\left\lvert - \frac{N_j}{N_{j+1}} m_j(E)+\frac 1{N_{j+1}}\sum_{k=1}^{j-1}\left(N_{k+1}-N_k\right)m_k(E)\right\rvert\leqslant 2\frac{N_j}{N_{j+1}}\leqslant j^{-1}$$ hence if $E$ is such that the sequence $\left(m_j(E)\right)_j$ does not converge to $\nu(E)$, the sequence $\left(\nu_{N_{j}}(E)\right)$ does not converge to $\nu(E)$.