Suppose $\{F_n\}$ is a sequence of probability measures on a compact set $X$ such that there exists a constant $M>0$ that for all $n$ and for any measurable sets $B$, we have
$$
F_n(B) \leq M \lambda(B),
$$
where $\lambda$ is the Lebesgue measure. Further, assume that $F_n$ converges weakly to a probability measure $F$.
It is obvious that for all $n$, $F_n$ is absolutely continuous wrt the Lebesgue measure.
Question: Is $F$ absolutely continuous and does it satisfy
$$
F(B) \leq M \lambda(B),
$$
for all measurable sets $B$?
Best Answer
Yes. Let $B$ be any measurable set. Let $\epsilon > 0$, then by the outer regularity of Lebesgue measure you can find an open set $U$ such that $B \subseteq U$ and $\lambda(U) < \lambda(B) + \epsilon$. Then by the portmanteau theorem we have $$ F(B) \leq F(U) \leq \liminf_{n \to \infty} F_n(U) \leq M\lambda(U) \leq M(\lambda(B) + \epsilon). $$ Since $\epsilon$ is arbitrary this implies the desired conclusion.