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Let $1\leq p<\infty.$ Let $(f_n)_n$ be a sequence in $L^{p}(\mathbb{R}^d)$ that converges to some $f\in L^{p}(\mathbb{R}^d)$, in $L^p$-norm. Prove that for every $\varepsilon>0$ we have $$l=\lim\limits_{n\rightarrow \infty}m\left(\left\{x\in\mathbb{R}^d:|f_n(x)-f(x)|>\varepsilon\right\}\right)=0,$$ where $m(E)$ denotes the Lebesgue measure of $E.$

My work so far:

Suppose for a contradiction that $l>0.$ Let $\varepsilon>0$ be given, and let $E_n=\left\{x\in\mathbb{R}^d:|f_n(x)-f(x)|>\varepsilon\right\}$. Since $f_n\rightarrow f$, in $L^p$-norm, there exists $N\in\mathbb{N}$ such that $\|f_n-f\|_{p}<\varepsilon\cdot m(E_N)^{1/p}$ for all $n>N,$ where $N$ is such that $m(E_N)>0,$ can be chosen since $l>0$. Then $$\left(\int_{E_N}\varepsilon^{p}\,dx\right)^{1/p}=\varepsilon\cdot m(E_N)^{1/p}<\left(\int_{\mathbb{R}^d}|f_n(x)-f(x)|^p\,dx\right)^{1/p}=\|f_n-f\|_p,$$ which is a contradiction, so we must have $l=0.$


Is my reasoning above correct? Any comments are very welcomed, be it about the correctness or the style of the proof, or both.

Thank you for your time, and appreciate any feedback.

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  • $\begingroup$ The fact that such $N$ exists is not that clear to me. You should give more arguments about this. $\endgroup$ – nicomezi Mar 2 '19 at 8:02
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    $\begingroup$ I don't understand why $\|f-f_n\|\leq \varepsilon m(E_N)$. Btw, if it would be true, roughly speaking your $\varepsilon $ would depend on $N$ which is not possible. $\endgroup$ – user649261 Mar 2 '19 at 8:03
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Honestly, it's just Markov inequality ! $$m\{|f_n-f|>\varepsilon \}\leq \frac{1}{\varepsilon ^p}\int|f_n-f|^p\to 0,$$

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  • $\begingroup$ If you don't mind me asking, was my approach correc? $\endgroup$ – Stackman Mar 2 '19 at 8:01
  • $\begingroup$ I answered in the comment of your question. @GabyAlfonso $\endgroup$ – user649261 Mar 2 '19 at 8:03
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    $\begingroup$ @stressedout: Let $E=\{|f_n-f|>\varepsilon \}$, then $\varepsilon \boldsymbol 1_E\leq |f_n-f|\boldsymbol 1_E\leq |f_n-f|$. Therefore $\varepsilon ^p\boldsymbol 1_{E}\leq |f_n-f|^p.$ Integrate both side and you'll get your result ;) $\endgroup$ – user649261 Mar 2 '19 at 8:06
  • $\begingroup$ Yup. I immediately realized that the proof is simple and proved it by myself. Thank you. $\endgroup$ – stressed out Mar 2 '19 at 8:16

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