Random variables converging to a density in mean

Question

Important edit: this question has been moved to Mathoverflow.

Let $X$ be an absolutely continuous r.v. with density $f$ which is continuous on $(0,\infty)$. Consider some a.s. decreasing sequence $Y_n$ bounded by $X$ such that $Y_n\searrow 0$ a.s. I stress that all variables are dependent (in a complicated way).

The question. I would like to know if the sequence $1_{\{x<X\leq x+Y_n\}}/Y_n$ converges in mean to $f(x)$ (i.e., $\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)\to f(x)$) under mild conditions (e.g. $\mathbb{E}(1/Y_n)<\infty$ for all $n$ or $X$ having bounded support).

Special cases. The following examples are not what I am after, but I include them to provide some intuition and perhaps clarity.

1. The independent case. (Edit: new version due to pre-kidney.) When $X$ and $Y_n$ are independent for each $n$, this is trivial. Indeed, the function $g_x:y\mapsto\mathbb{P}(x<X<x+y)/y$ is bounded and continuous on $[0,\infty)$ if we set $g_x(0)=g_x(0+)=f(x)$. Hence $$\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)= \mathbb{E}(\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n|Y_n))= \mathbb{E}(g_x(Y_n))\to f(x)$$ by the dominated convergence theorem (as $g_x(Y_n)\to f(x)$ a.s.).

2. The quotient can be sandwiched. Assume that some random variables $0<L_n$ and $U_n<1$ are independent of $X$ and satisfy $L_n\leq Y_n/X\leq U_n$ for all sufficiently large (deterministic) $n\in\mathbb{N}$ and $U_n/L_n\overset{L^1}{\to} 1$. In this case, we clearly have $U_n,L_n\overset{\mathbb{P}}{\to}0$ and, for sufficiently large $n$ $$1_{\{x<X\leq x/(1-L_n)\}}/(U_nX)\leq 1_{\{x<X\leq x+Y_n\}}/Y_n\leq 1_{\{x<X\leq x/(1-U_n)\}}/(L_nX).$$ Moreover, we may bound $$\mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-L_n)\}}}{U_nX}\Big|U_n,L_n\right) \geq \mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-L_n)\}}}{U_nx/(1-L_n)}\Big|U_n,L_n\right) =g_x\left(\frac{xL_n}{1-L_n}\right)\frac{L_n}{U_n}.$$ Since convergence in $L^1$ implies convergence in distribution, the continuous mapping theorem and the boundedness of $g_x$ imply that the last quantity converges in $L^1$ to $g_x(0)=f(x)$. Similarly, we bound $$\mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-U_n)\}}}{L_nX}\Big|U_n,L_n\right) \leq \mathbb{E}\left(\frac{1_{\{x<X\leq x/(1-U_n)\}}}{L_nx}\Big|U_n,L_n\right) =g_x\left(\frac{xU_n}{1-U_n}\right)\frac{U_n}{L_n}.$$ which again converges in $L^1$ to $g_x(0)=f(x)$. By sandwiching, we conclude that $\mathbb{E}(1_{\{x<X\leq x+Y_n\}}/Y_n)\to g_x(0)=f(x)$.

Flexibility of the result. It is not necessary that this exact result holds (for me). For instance, it would help if we have sufficient (and mild) conditions for: (I) $\mathbb{E}(1_{\{x<X\leq x+Y_n\}})/\mathbb{E}(Y_n)\to f(x)$ or (II) $\mathbb{E}[\mathbb{E}(1_{\{x<X\leq x+Y_n\}}|X)/\mathbb{E}(Y_n|X)]\to f(x)$ or something similar. If you have such a result, a reference would be much appreciated. Counterexamples are also welcome.

Answer 1

For $y>0$, let $g(x,y)=y^{-1}\mathbb P(x< X\leq x+y)$. By definition of the density, our hypotheses imply that $\lim_{y\to 0^+}g(x,y)=f(x)$.

Observe that (EDIT: as pointed out in the comments, this step actually is using independence of $X$ and $Y_n$) $\mathbb E(Y_n^{-1}1_{\{x< X\leq x+Y_n\}}\mid Y_n)=g(x,Y_n)$ a.s., and thus by continuity of $g$ and the fact that $Y_n\to 0$ a.s., it follows that $\lim_{n\to\infty} g(x,Y_n)=f(x)$ a.s.. Finally, we may remove the conditioning to obtain that $$ \lim_{n\to\infty}\mathbb E(Y_n^{-1}1_{\{x< X\leq x+Y_n\}})=\mathbb E\lim_{n\to\infty} g(x,Y_n)=\mathbb Ef(x)=f(x). $$