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

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 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. Feb 23, 2021 at 17:01

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 $$K$$.