TITLE: pointwise convergence and boundedness in norm imply weak convergence
QUESTION [10 upvotes]: I am contemplating over the following exercise (in which $E=[0,1]$):

Let $f_n$ be a sequence of functions in $L^p(E)$, $1<p<\infty$, which converge almost everywhere to a function $f$ in $L^p(E)$, and suppose that there is a constant $M$ such that $\|f_n\|_p\le M$ for all $n$. Then for each function $g$ in $L^q(E)$ (with $\frac1p+\frac1q=1$), we have 
  $$\int_E fg=\lim_{n\to\infty}\int f_n g.$$

If measure of $E$ is finite, one can make use of Egoroff's theorem and find a set $A$ of small enough measure such that $f_n$ converges uniformly to $f$ on $E\setminus A$, and $\int_A|g|^q<\epsilon'^q$. Thus the difference 
$$
\left|\int_E fg - \int_E f_ng\right|\le\int_E |(f-f_n)g|=\int_A |(f-f_n)g|+\int_{E\setminus A}|(f-f_n)g|
$$
$$
\le\text{[by Holder]}\le 2M\epsilon'+\|f-f_n\|_p\cdot\|g\|_q
$$
can be made less than any desired $\epsilon$ for big enough $n$.
My question is: does this result also hold true if the measure of $E$ is infinite?

REPLY [13 votes]: 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.