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I need to show that $f=\frac{1}{\sqrt x}$ is Lebesgue integrable on [0,1]. My attempt: I need to show $\sum_{m=n}^\infty \frac{m}{n} \mu(E_m^{(n)})$ converges absolutely $ \forall n$.

$\mu(E_m^{(n)})=\frac{n^2}{m^2}-\frac{n^2}{(m+1)^2}$

Then I fix $n=n_0$ and have this as my sum: $n_0 \sum_{m=n_0}^\infty \frac{2m+1}{m(m+1)^2}$ but I dont know how to proceed from this point.

Could you help me please.

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The sum is over terms that are $O(1/m^2)$ so it converges. –  Alex R. Oct 22 '12 at 5:04

1 Answer 1

$$\frac{2m+1}{m(m+1)^2}\leqslant\frac2{m^2}\qquad\qquad\qquad\sum_{m\geqslant1}\frac2{m^2}\lt\infty$$

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