Suppose we have the following sequence of random variables such that
\begin{align} Z_a&=E^2[V^2|aV+W]\\ \lim_{a \to 0} Z_a&= E^2[V^2] \quad (a.s.) \end{align} where $V$ and $W$ are independent and $W$ is not degenerate.
We want to find the limit \begin{align} \lim_{a \to 0} E[Z_a]=\lim_{a \to 0} E[E^2[V^2|aV+W] ] \end{align}
We can use Dominated Convergence Theorem to show the following. Since \begin{align} Z_a= E^2[V^2|aV+W] \le E[V^4|aV+W]=Y (\text{ by Jensen's inequality }) \end{align}
Moreover, $E[Y]=E[V^4]$ (** here assumed that $E[V^4]<\infty$ **). So by DCT we have that \begin{align} \lim_{a \to 0} E[Z_a]= E[ \lim_{a \to 0} Z_a]=E[E^2[V^2]]=E^2[V^2] \end{align}
My question is: Do we require the assumption that $E[V^4]<\infty$ to show the above result or can we find the limit with out this assumption?
This question is somewhat related to this question.
Thanks for any help.
Edit (by extrapolating suggestion A.S. )
Suppose $f_W(w)$ is bounded, that is $f_W(w) \le a$ then \begin{align} E[V^2|aV+W=x]&= \int v^2 \frac{f_{aV+W|V}(x|v)}{f_{aV+W}(x)} dV(v)=\int v^2 \frac{f_{aV+W}(x-av)}{f_{aV+W}(x)} dV(v) \le \int v^2 \frac{c}{f_{aV+W}(x)} dV(v)\\ & \frac{c}{f_{aV+W}(x)} \int v^2 dV(v)=\frac{c}{f_{aV+W}(x)} E[V^2]\\ \end{align}
So, now \begin{align} E[ Z_a] \le E \left[ c^2 E[V^2] \frac{1}{(f_{aV+W}(aV+W))^2} \right]=c^2 E[V^2] E \left[ \frac{1}{(f_{aV+W}(aV+W))^2} \right] \end{align}
But what to do with the term $E \left[ \frac{1}{(f_{aV+W}(aV+W))^2} \right]$??