Mathematics Asked on January 3, 2022

Question: Suppose $f_n, g_n:[0,1]rightarrowmathbb{R}$ are measurable functions such that $f_nrightarrow 0$ a.e. on $[0,1]$ and $sup_nint_{[0,1]}|g_n|dx<infty$.

- 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 $nrightarrowinfty$.
- Prove that for any such sequences $f_n$ and $g_n$, and every $epsilon>0$, there exists a measurable set $Esubset[0,1]$ such that $m(E)>1-epsilon$ and $int_Ef_n g_ndxrightarrow 0$.

My thoughts: I was thinking of doing something like $f_n=nchi_{(0,frac{1}{n}]}$, 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.

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 $k=sup_nint_0^1|g_n|$. By Egorov's Theorem there exists $Esubset[0,1]$ with $m(E)>1-varepsilon$ and $f_nto0$ uniformly on $E$. . So there exists $n_0$ such that, for all $n>n_0$, we have $|f_n|leqvarepsilon/k$. Then $$ int_E|f_ng_n|leqfracvarepsilon k,int_E|g_n|leqvarepsilon. $$

Answered by Martin Argerami on January 3, 2022

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

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

Answered by cha21 on January 3, 2022

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