Why does this statement hold in the infinite case?

Question

I have been reading a problem posted here and could not really understand the answer. The statement is as follows if $f_n \rightarrow f$ a.e. and they are both in $L^p(E)$ with $1<p<\infty$ and $\|f_n\|_p \leq M < \infty$ then for every $g \in L^q(E)$ where $1/p+1/q=1$ we have $$\lim_{n \rightarrow \infty} \int_E f_n g = \int_E fg.$$ The case where $E$ is finite is not too bad just some clever usage of Egorov's theorem, etc. Now for the infinite case the accepted answer seems to say that you just simply write $E = \bigcup S_n$ where $S_n = \{x: |g(x)| > 1/n\}.$ Take $E = \mathbb{R}$ for example.

Now I have two problems with this:

1. If $g(x) = 0$ for some $x \in E$, then the point $x$ would not be in any of the $S_n$'s would it?

2. Another problem, which I think is more important, is the claim that $\int_{E \setminus S_n} |g|^q < \epsilon$. Now I know we can make this true since $|g|^q$ is integrable and so this would hold as long as $m(E\setminus S_n) < \delta$. However I think what is being hinting at is using continuity from below of the Lebesgue measure, that is $$\lim_{n \rightarrow \infty} |S_n| = |E|.$$ But $|E| = \infty$ so how could we possibly make $|E \setminus S_n| = |E| - |S_n| < \epsilon?$

Anyways, I am not sure if this approach works, but most likely I am misunderstanding something here ...

Is there a way to prove the infinite case not using this line of thought?

Any help is much appreciated.

Krull.

Answer 1

For your first question, note that \begin{align*} \lim_{n}\int f_{n}g=\int fg \end{align*} is equivalent to \begin{align*} \lim_{n}\int_{(g\ne 0)} f_{n}g=\int_{(g\ne 0)}fg, \end{align*} so one can concentrate on all those $S_{N}$.

EDIT:

Note that $C_{0}$, the set of continuous functions with compact support is dense in $L^{q}$.

Let $\epsilon>0$ be given.

For $g\in L^{q}$, choose a $\varphi\in C_{0}$ such that $\|g-\varphi\|_{L^{q}}<\epsilon$.

So we have proved that \begin{align*} \lim_{n}\int|f_{n}-f||\varphi|=0 \end{align*} and hence there is some $N$ such that \begin{align*} \int|f_{n}-f||\varphi|<\epsilon,~~~~n\geq N, \end{align*} for all such $n$, we have \begin{align*} \int|f_{n}-f||g|&\leq\int|f_{n}-f||g-\varphi|+\int|f_{n}-f||\varphi|\\ &\leq\|f_{n}-f\|_{L^{p}}\|g-\varphi\|_{L^{q}}+\epsilon\\ &\leq 2M\epsilon+\epsilon. \end{align*}

Answer 2

Here is another approach for the second question.

We assume that $E=\bigcup_{N}Q_{N}$ for disjoint finite measure $Q_{N}$, or else I think there is some counterexample for non-$\sigma$-finite $E$.

We write \begin{align*} |g|=\sum_{N}|g|\chi_{Q_{N}}, \end{align*} and we see that \begin{align*} \lim_{n}\int|f_{n}-f||g|&=\lim_{n}\sum_{N}\int|f_{n}-f||g|\chi_{Q_{N}}. \end{align*} The concern is to swipe $\lim_{n}$ with the sum $\sum_{N}$, this is a matter of Lebesgue Dominated Convergence Theorem once we manage to see that \begin{align*} \int|f_{n}-f||g|\chi_{Q_{N}}\leq\|f_{n}-f\|_{L^{p}}\|g\chi_{Q_{N}}\|_{L^{q}}\leq 2M\|g\chi_{Q_{N}}\|_{L^{q}} \end{align*} and that \begin{align*} \sum_{N}\|g\chi_{Q_{N}}\|_{L^{q}}=\|g\|_{L^{q}}<\infty. \end{align*} We are justified to do the swiping, hence \begin{align*} \lim_{n}\int|f_{n}-f||g|=\sum_{N}\lim_{n}\int|f_{n}-f||g|\chi_{Q_{N}}=0. \end{align*}