pointwise convergence and boundedness in norm imply weak convergence

Solution 1:

Just want to add another approach to this problem.

To show $\lim \int f_n g = \int fg$ for each $g\in L^q$, if $f_n$ are bounded, it suffices to show $\lim \int f_n \phi = \int f\phi$ for $\phi$ in a dense subset of $L^q$, say the space of simple functions in $L^q$ with compact support. Then your argument would apply perfectly.

Solution 2:

Yes: consider for each integer $N$ the sets $S_N:=\{x,|g(x)|>N^{-1}\}$. Since $|g|^q$ is integrable, these sets have finite measure, and we are reduced to show the result when $E=\bigcup_NS_N$.

Then we use a $2\varepsilon$-argument: fix $\varepsilon>0$; we can find $N$ such that $\int_{E\setminus S_N}|g|^q<\varepsilon$. Then we do the same proof as the case $E$ of finite measure.