Suppose we have a sequence of random variables $f_n:\Omega\rightarrow\mathbb{R}$ converging almost surely to some random variable $f:\Omega\rightarrow\mathbb{R}$. Suppose we know that the law of each $f_n$ has a continuous density function $\phi_n:\mathbb{R}\rightarrow\mathbb{R}$ and $f$ has a continuous density $\phi:\mathbb{R}\rightarrow\mathbb{R}$.
I know that almost sure convergence of the random variables implies weak convergence of the laws. Can we use the extra assumptions here to conclude that we have a stronger form of convergence? For example it seems like we should have $\|\phi_n-\phi\|_{L^1}\rightarrow 0$.
Best Answer
Counterexample: Consider the prbability space $(\Omega,\mathcal{F},\mathbb{P})=([0,\pi],\mathcal{B}([0,\pi]),\frac1\pi\mathrm{Leb})$ and random variables $X_n\colon[0,\pi]\to[0,\pi]$ defined by its inverse $$ X_n^{-1}(x)=x+\frac{\sin(2nx)}{2 n} $$ Clearly $X_n$ converges uniformly (hence also almost surely) to the identity $X_\infty\colon[0,\pi]\to[0,\pi]\subset\mathbb{R}$. But the probability density functions $\phi_n(x)=\frac2\pi\cos^2(nx)$ does not converge in $L^1$ to $\phi_\infty=\frac1\pi 1_{[0,\pi]}$. Indeed, $\lVert\phi_n-\phi_\infty\rVert_{L^1}=2$ for all $n$.