I feel like it's a well established fact that if the $L^1$ norm of an integrable function is zero then, it must be zero almost everywhere, but what's the proof?
$\|f\|_1=0$ if and only if $f=0$ almost everywhere
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lebesgue-integral
1 Answers
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Consider the measurable sets $E_n = \{x: |f(x)| \ge 1/n \}$. Then $\bigcup_{n=1}^\infty E_n = \{x: f(x) \ne 0\}$. If it's not the case that $f= 0$ almost everywhere, this has nonzero measure, so by countable additivity some $E_n$ has nonzero measure. If $m(E_n) > 0$, then $\|f \| \ge \int_{E_n} |f| \ge m(E_n)/n > 0$.