Clarification: Here $\mu$ being absolutely continuous means being absolutely continuous with respect to the Lebesgue measure $dx$: $\mu(A)=\int_A fdx$ for some $f$ for all Lebesgue measurable $A$. Having bounded density means the density functions of these probability measures are uniformly bounded by a constant. Also, closed means closed under weak topology on the space of probability measures of $\mathbb{R}$, $\mu_n$ converge to $\mu$ if and only if $\int_\mathbb{R} fd\mu_n\to\int_\mathbb{R} fd\mu$ for all bounded continuous $f$.
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2$\begingroup$ Is the uniform bound fixed for the whole collection of probabilities? As in, you are considering the set of $f(x)dx$ for $f$ bounded by, say, 1? If not, the Gaussian distributions $\mathcal N(0,1/n^2)$ are a counterexample. $\endgroup$– Pierre PCCommented Feb 23, 2021 at 17:01
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The answer is yes. Indeed, a probability measure $\mu$ over $\mathbb R$ has a density bounded by a real $K>0$ iff the cdf of $\mu$ is $K$-Lipschitz, that is, Lipschitz with the Lipschitz constant $K$.
So, you have a sequence $(\mu_n)$ of probability measures over $\mathbb R$ with $K$-Lipschitz cdf's $F_n$ converging to the cdf $F$ of a probability measure $\mu$ at all points of continuity of $F$. Since the set of all points of continuity of $F$ is dense in $\mathbb R$, we conclude that $F$ is $K$-Lipschitz. So, $\mu$ has a density bounded by $K$.
answered Feb 23, 2021 at 17:08
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$\begingroup$ Thanks! This is exactly what I was looking for. $\endgroup$– kid111Commented Feb 24, 2021 at 3:35