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

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.

• The fact that such $N$ exists is not that clear to me. You should give more arguments about this. – nicomezi Mar 2 '19 at 8:02
• 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. – user649261 Mar 2 '19 at 8:03

Honestly, it's just Markov inequality ! $$m\{|f_n-f|>\varepsilon \}\leq \frac{1}{\varepsilon ^p}\int|f_n-f|^p\to 0,$$
• @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 ;) – user649261 Mar 2 '19 at 8:06