Let $X,Y,Z$ be three positive random variables. I want to find the following limit \begin{align} \lim_{\epsilon \to 0} E \left[ \frac{Y}{(1-\epsilon)X+\epsilon Z} \right]. \end{align}
I was wondering if I there is an issue if I apply montone convergence theorem together with conditioning as follows:
\begin{align} \lim_{\epsilon \to 0} E \left[ \frac{Y}{(1-\epsilon)X+\epsilon Z} \right]=\lim_{\epsilon \to 0} E \left[ \frac{Y}{(1-\epsilon)X+\epsilon Z} \Big| X \ge Z\right] P[X \ge Z] + \lim_{\epsilon \to 0} E \left[ \frac{Y}{(1-\epsilon)X+\epsilon Z} \Big| X < Z\right] P[X < Z]. \end{align}
Under the condition $X \ge Z$ we have that \begin{align} W_\epsilon=\frac{Y}{(1-\epsilon)X+\epsilon Z}, \end{align} is monotone increasing and under condition that $X \le Z$ we have that $W_\epsilon$ is motone decreasing.
So using monotone convergence theorem on both of these conditional expectations we have that the limit is \begin{align} \lim_{\epsilon \to 0} E \left[ \frac{Y}{(1-\epsilon)X+\epsilon Z} \right]=E \left[ \frac{Y}{X} \right]. \end{align}
Is this a correct proof? Did I miss anything?
The result is true as stated, but requires a more elaborate proof. Lemma If $\{f_n\}$ is decreasing sequence of measurable functions with a.e. limit f and if at least one $f_n$ is integrable then $\int f_n d\mu \to \int f d\mu$. This follows by applying monotone convergence theorem to $\{f_1-f_n\}$. The argument of the OP along with this lemma gives the result easily when $E\frac Y X <\infty$. To prove the result when $E\frac Y X = \infty$ fix a positive integer N and let $Y_1=Y$ when $Y \leq NX$ and 0 otherwise. Note that $0 \leq Y_1 \leq Y$. Hence $E \frac Y {(1-\epsilon)X+\epsilon Y}$ $\geq E \frac {Y_1} {(1-\epsilon)X+\epsilon Y_1}$. ( This requires a little argument, but it is easy). Noting that $E\frac {Y_1} X <\infty$ we can apply the previous result to get $liminf E \frac Y {(1-\epsilon)X+\epsilon Y} \geq E\frac {Y_1} X$. Finally $E\frac {Y_1} X \to \infty$ as $N \to \infty$. The rest is clear.