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Let $f_n\in L^1[0,1]$ be positive functions such that:

$$\int_0^1 f_n=1$$

$$\int_{\frac{1}{n}}^{1}f_n<\frac{1}{n}$$

for all n, prove:

$$\int_0^1 \sup_{n\geq 1}f_n =\infty$$

I don't know how to handle this, who can give me some suggestions?

Thank all of you!

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Hint: If $\int_0^1 g<\infty$, then $\int_0^{\frac1n}g\to0$ as $n\to\infty$.

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  • $\begingroup$ Thank you very much. I know how to prove now. $\endgroup$ – Kira Yamato Nov 22 '14 at 1:01
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Hint: Devide $\int_0^1 sup_{n\in \mathbb N} f_n $ into $n$ integrals. In each integral (or interval), find $f_n$ which it's integral would be big enough.

It should probably work if you devided into two and used $\int_0^{1/n} f_n>1-1/n \rightarrow 1$.

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