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

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.

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 $$nrightarrowinfty$$.
2. 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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