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.
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.
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
$$\|f\|_p^p = p \int_0^\infty x^{p - 1} \mu\{z : |f(z)| > x\} \, dx$$
So,
\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 $$\int_{2}^\infty \mu \{x \in [0,1] : |f(x)| > t \} \, dt \leq \int_{2}^\infty \frac{1}{t \log^2 t} \, dt = \frac{1}{\log(2)}.$$