Let $X,Y$ be nonzero independent random variables, and $X$ has zero expectation. Let the distributions of $X,Y$ be $\mu,\nu$ respectively. Problem gives a condition that $$0 < \lim_{x\to 0+} \frac{\nu(-x,x)}{x} = a< \infty$$ Define $Z_n = X/(|Y| + 1/n)$ and let $Z = \lim_n Z_n$. The problem is to compare the expectations of $Z_n$'s and $Z$.
For $Z_n$, since $E\frac{1}{|Y| + 1/n} \leq E[n] = n < \infty$, we have $$E \frac{X}{|Y| + 1/n} = EX \cdot E\frac{1}{|Y|+1/n} = 0$$ since both expectations are finite. However, in the case of $Z$, We can't say that $EZ = 0$ since $E\frac{1}{|Y|}$ need not be finite. My try is to use transformation theorem and Fubini's theorem: $$E \frac{X}{|Y|} = \int_{\mathbb R^2} \frac{x}{|y|} d (\mu \otimes\nu) = \int_{\mathbb R} \int_{\mathbb R} \frac{x}{|y|} d\mu(x) d\nu(y)$$ but we cannot further separate the integrals since they are not guaranteed to be finite. How can I deal with such a situation? What can be the meaning of the condition $0 < \lim_{x\to 0+} \frac{\nu(-x,x)}{x} =a< \infty$?
Thanks in advance for any form of help, hint, or solution.
For any $0<\varepsilon<1$, \begin{align*} \Bbb E\,{\frac1{|Y|}} &=\int_0^\infty\frac{\nu(-x,x)}{x^2}\,\mathrm dx\\[.4em] &\ge\int_{\varepsilon^2}^\varepsilon\frac{\nu(-x,x)}{x^2}\,\mathrm dx\\[.4em] &\ge\left(\inf_{\varepsilon^2\le x\le\varepsilon}\frac{\nu(-x,x)}x\right)\int_{\varepsilon^2}^\varepsilon\frac{\mathrm dx}x\\[.4em] &\ge\left(\inf_{0<x\le\varepsilon}\frac{\nu(-x,x)}x\right)\log\left(\frac1{\varepsilon}\right), \end{align*} so $$\Bbb E\,\frac1{|Y|}\ge\underbrace{\liminf_{x\to0^+}\frac{\nu(-x,x)}x}_{{}>0}\cdot\left[\lim_{x\to0^+}\log\left(\frac1x\right)\right]=\infty.$$ This is equivalent to $\Bbb E\,|Z|=\infty$ because $\Bbb E\,|X|\in(0,\infty)$ ($X$ is nonzero and has finite expectation).
In fact, one can see that a necessary and sufficient condition for $\Bbb E\,\frac1{|Y|}<\infty$ is that $\frac{\nu(-x,x)}{x^2}$ be integrable at $0^+$, so that $\nu(-x,x)$ decays slightly faster than $x$ (e.g., $\nu(-x,x)=O(x/{\log^2x})$ would work).