Suppose $(X,\mu)$ is a probability space, $W\in L^1(X)$, $V\in L^\infty(X)$, and $V_n\to V$ in $L^2(X)$ (in my situation $V_n$ is the partial Fourier sum and so the $L^2(X)$ convergence is automatic).
Can we say that $\int_X WVd\mu= \lim_{n\to\infty}\int_X WV_nd\mu$?
I think so, but I am having trouble showing this. If $W$ is square integrable then the result follows from Cauchy-Schwarz. But of course this argument does not work for less regular $W$. I have considered dominated convergence theorem and it's variation, but finding a dominating function is eluding me. Could anyone help out? Both suggestions and solutions are welcome.
Thanks in advance!
Consider the probability space $X=[0,1]$ with uniform probability measure $\mu$. We let $V=0$ and $W$ be defined by $$ W(x) :=\sum_{n=1}^\infty \frac {2^n}{n^2} \chi_{[2^{-n}, 2^{-n+1}]}, $$ it is not hard to verify that $W\in L^1(X,\mu)$.
Now, let the sequence $V_n$ be defined as $$ V_n(x):= 2^{n/4} \chi_{[2^{-n}, 2^{-n+1}]}, $$ it is also easy to compute that $V_n\to 0$ in $L^2(X,\mu)$.
However, we have $$ \int_X WV\,d\mu = 0 \ne \infty = \lim_{n\to\infty} \frac 1{n^2}\cdot \frac{2^{5n/4}}{2^n}= \lim_{n\to\infty}\int_X WV_n\,d\mu. $$