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Let $f\in L^1(\mathbb R)$ s.t. $$\int_{\mathbb R}|f|=0\implies f=0\ a.e.$$

My attempt

Suppose $f$ continuous and that there is $y$ s.t. $|f(y)|\neq 0$. In particular, there is $\delta >0$ s.t. $f(x)\neq 0$ for all $x\in [y-\delta ,y+\delta ]$. By continuity, there is $m>0$ s.t. $|f(x)|\geq m$ for all $x\in [y-\delta ,y+\delta ]$. Therefore, $0<2m\delta\leq \int_{[y-\delta ,y+\delta ]}|f(x)|\leq \int_{\mathbb R}|f|,$ which is a contradiction. Therefore $f=0$ everywhere.

Since $f\in L^1$. There is a sequence of continuous function s.t. $$\|f_n-f\|_{L^1}\to 0.$$ In particular, $$\int_{\mathbb R}|f_n|\to \int_{\mathbb R}|f|=0.$$

How can I prove that $f_n=0$ for all $n$ ?

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    $\begingroup$ Why would $f$ be continuous? How can you suppose it? Suppose instead $f > 0$ on non null set. $\endgroup$ – dEmigOd Feb 16 at 10:19
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    $\begingroup$ @dEmigOd. If you continue reading the proof you will see that he first considers the case where $f$ is continuous, and then approximates a general $f \in L^1$ with continuous functions. $\endgroup$ – md2perpe Feb 16 at 10:51
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There is no reason to have $f_n=0$ for all $n$... Your proof would be fine if $$\left\{f\in L^1(\mathbb R)\mid \int_{\mathbb R}|f|=0\right\},$$ would be a dense subspace of $L^1(\mathbb R)$. Since it's not the case, your proof can't work.

For a proof, let $E=\{x\mid |f(x)|>0\}$. Suppose that $m(E)>0$. Let $$E_n=\left\{x\mid |f(x)|\geq\frac{1}{n}\right\}.$$ Then, $$\frac{1}{n}\boldsymbol 1_{E_n}<|f|\boldsymbol 1_{E_n} \leq |f|,$$ and thus, $$\frac{1}{n}m(E_n)\leq \int_{\mathbb R}|f|=0.$$ Therefore, $m(E_n)=0$ for all $n$. Since $\lim_{n\to \infty }m(E_n)=m(E)$, you get a contradiction.

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