Let $X\_n, X$ be $[0, 1]$-valued random variables whose laws are absolutely continuous with respect to Lebesgue measure. Suppose $X\_n \to X$ a.s. Does this imply that the pdfs of $X\_n$ converge to that of $X$ in some suitable sense?


For concreteness, the three I have in mind are convergence a.e., convergence in $L^p$, and convergence in measure; where the measure theoretic statements are with respect to the Lebesgue measure. However I suspect none of these are true. On the other hand, it would seem unusual for there to be no convergence at all on the pdfs. What is the right type of convergence to be looking at?



Almost sure convergence vs convergence of probability density functions
The answer is no. Indeed, for natural $n$ let
$$p\_n(x):=1+\tfrac12\,\text{sign}\,\sin(2\pi nx)$$
if $x\in[0,1]$, with $p\_n(x):=0$ if $x\notin[0,1]$. Then $p\_n$ is a pdf, with the corresponding cdf $F\_n$, so that
$$F\_n(x)=\int\_{-\infty}^x p\_n(t)\,dt$$
for all real $x$. The cdf $F\_n$ is continuously and strictly increasing on $[0,1]$, with $F\_n(0)=0$ and $F\_n(1)=1$. Hence, for each $u\in(0,1)$ there is a unique root $x=F\_n^{-1}(u)$ of the equation $F\_n(x)=u$. Moreover, the function $F\_n^{-1}\colon(0,1)\to\mathbb R$ is continuous.


Further, $F\_n(x)\to x$ for all $x\in[0,1]$ and hence $F\_n^{-1}(u)\to u$ for all $u\in(0,1)$ (as $n\to\infty$).


Let now $X\_n:=F\_n^{-1}(U)$ and $X:=U$, where $U$ is any random variable uniformly distributed on $(0,1)$. Then $X\_n\to X$ almost surely. Also, $F\_n$ is the cdf of $X\_n$ and hence $p\_n$ is the pdf of $X\_n$.


However, the pdf $p\_n$ of $X\_n$ does not converge to the pdf (say $p$) of $X$ even in measure -- because otherwise, by dominated convergence, $p\_n$ would converge to $p$ in $L^1$, whereas the $L^1$ norm of $p\_n-p$ is
$$\int\_0^1|p\_n(x)-1|\,dx=\frac12\not\to0.$$




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On a positive note, the almost sure convergence of $X\_n$ to $X$ will of course imply the convergence of $X\_n$ to $X$ in distribution, so that the pdf $p\_n$ of $X\_n$ will converge to the pdf $p$ of $X$ in the weak sense that
$$\int\_{\mathbb R}f(x)p\_n(x)\,dx\to\int\_{\mathbb R}f(x)p(x)\,dx$$
for each bounded continuous function $f\colon\mathbb R\to\mathbb R$.


