Two sequences $f_n$ and $g_n$ such that $\int_{[0,1]}f_n g_n$ does not go to $0$ as $n\rightarrow\infty$, with these conditions on $f_n$ and $g_n$

Question

Question: Suppose $f_n, g_n:[0,1]\rightarrow\mathbb{R}$ are measurable functions such that $f_n\rightarrow 0$ a.e. on $[0,1]$ and $\sup_n\int_{[0,1]}|g_n|dx<\infty$.

1. Give an example of two sequences $f_n$ and $g_n$ such that $\int_{[0,1]}f_n g_n$ does not go to $0$ as $n\rightarrow\infty$.
2. Prove that for any such sequences $f_n$ and $g_n$, and every $\epsilon>0$, there exists a measurable set $E\subset[0,1]$ such that $m(E)>1-\epsilon$ and $\int_Ef_n g_ndx\rightarrow 0$.

My thoughts: I was thinking of doing something like $f_n=n\chi_{(0,\frac{1}{n}]}$, which I believe would converge pointwise to $1$ a.e...I am just having a hard time trying to think of a $g_n$ that would work such that the integral of their product over $[0,1]$ wouldn't go to $0$.... For the second question, I immediately was thinking Egorov, but I haven't quite been able to figure out how to use it here.

Any suggestions, ideas, etc. are appreciated! Thank you.

Answer 1

Your functions $f_n$ work fine if you take $g_n=1$ for all $n$, since $$ \int_0^1f_n=1 $$ for all $n$.

Given any such pair $\{f_n\}$, $\{g_n\}$, and $\varepsilon>0$, let $k=\sup_n\int_0^1|g_n|$. By Egorov's Theorem there exists $E\subset[0,1]$ with $m(E)>1-\varepsilon$ and $f_n\to0$ uniformly on $E$. . So there exists $n_0$ such that, for all $n>n_0$, we have $|f_n|\leq\varepsilon/k$. Then $$ \int_E|f_ng_n|\leq\frac\varepsilon k\,\int_E|g_n|\leq\varepsilon. $$

Answer 2

For the first part you can just take $g_n = 1$

For the second part fix $\epsilon > 0$. For every $n > 0$, let $A_n \subset [0,1]$ be a set with measure greater than $1 - \frac{\epsilon}{2^n}$ such that there exists an $N \in \mathbb{Z}_+$, so that for $m \geq N$, if $x \in A_n$, $|f_m(x)| < \frac{1}{2^n}$. Take $$E = \bigcap_n A_n$$ Then

$$|\int_E f_ng_n| \leq \sup_{x \in E} |f_n(x)| \sup_m\int_{[0,1]} |g_m| \rightarrow 0$$ as $n \rightarrow \infty$