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Let $f:[0,1] \rightarrow [0,\infty)$ be a measurable function such that:

$\mu (\{x \in [0,1]: f(x) > t \}) \leq \frac{1}{t(ln(t))^{2}}$

holds for each $t>3$.

Show $f$ is an integrable map.

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Looks like homework. What you have tried? – Aryabhata Nov 13 '10 at 15:25
It is not, just practice for the final. That's the problem I don't even know where to start. Any hint? – student Nov 13 '10 at 15:27
Show that $\int_0^1f(x)\;dx = \int_0^\infty \mu (\{x \in [0,1]: f(x) > t \})\;dt$. – Stefan Nov 13 '10 at 16:15

For the sake of having less open questions and the fact that the final is probably already done I will show that if $f \in L^p(X,\mu)$, $0 < p < \infty$ then

$$\|f\|_p^p = p \int_0^\infty x^{p - 1} \mu\{z : |f(z)| > x\} \, dx$$


\begin{align*} p \int_0^\infty &x^{p - 1} \mu\{z : |f(z)| > x\} \, dx\\ &= p \int_0^\infty \int_X 1_{{z : |f(z)| > x}} \, d\mu(z) \, dx \text{ writing the measure as an integral.}\\ &= \int_X \int_0^{|f(z)|} p x^{p - 1} \, dx \, d\mu(z) \text{ Fubini,}\\ &= \int_X |f(x)|^p \, d\mu(x) \end{align*}

So, you have $p = 1$, so $$\int_{2}^\infty \mu \{x \in [0,1] : |f(x)| > t \} \, dt \leq \int_{2}^\infty \frac{1}{t \log^2 t} \, dt = \frac{1}{\log(2)}.$$

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Nice job. This is probably dumb but why did you integrate starting from 2? – user1736 Nov 19 '10 at 22:20
@user1736: Well, the thing makes sense from two on, not three. I just took the most acceptable one because it is not given what it does for $t \leq 3$. – Jonas Teuwen Nov 27 '10 at 19:22

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