# pointwise convergence and boundedness in norm imply weak convergence

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?

• Are you sure that your proof when $E$ has finite measure works? What about the term $||f-f_{n}||_{p}.||g||_{q}$? How do you intend to remove that? – Riju Jul 11 '17 at 18:58
• @Riju $f_n$ converges uniformly to $f$ on a set of finite measure, hence it converges in $L_p$ norm as well. So the term $\|f-f_n\|_p$ can be made arbitrarily small, while $\|g\|_q$ is bounded. – mathreader Jul 11 '17 at 23:03

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.
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.