Proving that $g_nf_n$ converges to $0$ in measure on $[0,1]$

Question: Let $f_n,g_n:[0,1]rightarrow [0,infty)$ be measurable functions. Assume $f_nrightarrow 0$ in measure on $[0,1]$, and that $int g_ndx<1$ for all $ninmathbb{N}$. Prove that $g_nf_nrightarrow 0$ in measure on $[0,1]$

My thoughts: So, we are trying to show that for all $epsilon>0$, there exists $N$ such that $forall n>N$, we have $m{|g_nf_n-gf|>epsilon}<epsilon$. So, I was going to try and show convergence pointwise a.e. to then imply convergence in measure, but that fell apart. So, would I go about this by setting up integrals and splitting the integral bounds?

Any help, suggestions, tips, etc. are (as always!) greatly appreciated! Thank you.

Mathematics Asked by User7238 on November 14, 2021

2 Answers

2 Answers

$$ begin{align} lambda(|f_ng_n|>varepsilon)&=lambda(|f_ng_n|>varepsilon,|f_n|>varepsilon^2)+lambda(|f_ng_n|>varepsilon,|f_n|leqvarepsilon^2)\ &leq lambda(|f_n|>varepsilon^2) + lambda(varepsilon^2|g_n|>varepsilon)\ &leq lambda(|f_n|>varepsilon^2) + lambda(|g_n|>tfrac{1}{varepsilon})leq lambda(|f_n|>varepsilon^2) + varepsilonint|g_n|\ &leq lambda(|f_n|>varepsilon^2) + varepsilon end{align} $$

Then for all $n$ large enough, say $ngeq N_varepsilon$, $$lambda(|f_ng_n|>varepsilon)leq 2varepsilon$$ This implies that $g_nf_n$ converges to $0$ in measure. To see this, let $delta>0$. Choose $varepsilon<delta$ and let $N_varepsilon$ be as above. Then $lambda(|f_ng_n|>delta)leq lambda(|f_ng_n|>varepsilon)leq2varepsilon$ for all $nleq ngeq M_varepsilon$. This shows that $lim_{nrightarrowinfty}lambda(|f_ng_n|>delta)=0$.

Answered by Oliver Diaz on November 14, 2021

Since convergence in measure is equivalent the fact that every subsequence has a further subsequence converging almost everywhere we can reduce the proof to the case where $f_n$ tends to $0$ almost everywhere.

With this change the result can be proved easily using Egoroff's Theorem. Choose $E$ such that $f_n to 0$ unifromly on $E$ and $mu (E^{c}) <epsilon$. Choose $n_0$ such that $ f_n(x) <epsilon ^{2}$ for all $x in E$ for all $n geq n_0$. Now $mu (f_ng_n >epsilon) leq epsilon + mu (E cap (f_ng_n >epsilon))leq epsilon+mu (g_n >frac 1 {epsilon}) leq epsilon+epsilon int g_n dmu<2epsilon$.

Answered by Kavi Rama Murthy on November 14, 2021

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