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The hint is to use the fact that $\int_{X}|f|d\mu<\infty$.

$E:=\{x\in X\mid f(x)\neq 0\}=\{x\in X\mid |f(x)|\neq 0\}$

$\int_{X}|f|d\mu:=sup\{\int_{X}ud\mu\mid 0\leq u\leq |f|$ and $u$ is a simple function$\}$.

I'm not sure what to do with these information.

Sid Caroline
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    That "hint" is not really a hint. Its only message is: "don't forget that $f$ is integrable here." – Vera Jun 16 '18 at 12:30

1 Answers1

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Let $E_n:=\{x\in X\mid |f(x)|\geq\frac1n\}$ for positive integer $n$.

Then $E=\bigcup_{n=1}^{\infty}E_n$ so it is enough to prove that $\mu(E_n)<\infty$ for every $n$.

Now observe that $1_{E_n}(x)\leq n|f(x)|$ for every $x\in X$ so that: $$\mu(E_n)=\int1_{E_n}d\mu\leq\int n|f|d\mu=n\int |f|d\mu<\infty$$

Vera
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