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 $f$.



Is the set of probability measures on $\mathbb{R}$ absolutely continuous with bounded density a closed subset?
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 $K$.


