If $f_n\rightarrow f$ in $L^p$-norm, $1\leq p<\infty$, then $\lim\limits_{n\rightarrow\infty}m(\{x\in\mathbb{R}^d:|f_n(x)-f(x)|>\varepsilon\})=0$

Question

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 $l=0.$

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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.

Answer 1

Honestly, it's just Markov inequality ! $$m\{|f_n-f|>\varepsilon \}\leq \frac{1}{\varepsilon ^p}\int|f_n-f|^p\to 0,$$