This is probably a trivial exercise, but I'm having trouble with deriving a formal proof.
Let $X$ be a r.v. with values in $[0,\infty)$. Show that $$ \lim_{\epsilon \downarrow 0} \frac{1}{\epsilon} \int_{X\leq \epsilon}X=0$$.
So far I've thought of reformulating this as$$ \mathbb{E}[\frac{X}{\epsilon_n}1_{[0,\epsilon_n]}]\rightarrow0$$ where $\epsilon_n \downarrow 0$.
I don't know how to proceed.
With Lebesgue DCT: consider any positive sequence $(\epsilon_n)$ that goes to $0$ and let $Y_n:=1_{X\leq \epsilon_n}\frac{X}{\epsilon_n}$.
Then $0\leq Y_n\leq 1$ a.s. and $Y_n\to 0$, hence $E(Y_n)\to 0$.
Without DCT: note that $E(1_{X\leq \epsilon_n}\frac{X}{\epsilon_n})\leq E(1_{0<X\leq\epsilon_n}) = F_X(\epsilon_n)-F_X(0)$ where $F_X$ denotes the cdf of $X$. Since $F_X$ is right-continuous, we obtain $\lim_n F_X(\epsilon_n)-F_X(0) = 0$, hence the result.
