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Let $(\Omega, A, P)$ be a probability space, I have to show that for a measurable function $u$:

$u\in \mathcal{L}^1(P) \iff \sum_{j=0}^{\infty} P(\{|u|\geq j\})<\infty$

I wanted to apply the "Markov's inequality", to each the single term

$$P(\{|u|\geq j\})$$ and then sum them up, mind that $j\geq 0$ while the requisite for applying the Markov's inequality is that $j>0$.

But even if we ignore the latter (because $P(\{|u|\geq 0\})=P(\Omega)$) we are left with something like

$$\sum_{j=0}^{\infty} P(\{|u|\geq j\})\leq 1+\sum_{j=1}^{\infty} \frac{1}{j} \int |u |dP$$ which is inconclusive because $\sum_{j=0}^{\infty} \frac{1}{j}$ does not converge.

If I knew that $ \int |u |^2 dP<\infty$ I would substitute the $(1/j)$ by $(1/j)^2$ but since it's not the case I am a little bit confused.

I would really appreciate you give me some hint instead of the complete solution since I would like to solve it by myself.

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  • $\begingroup$ Take e.g. a look at this question or one of its many many duplicates. $\endgroup$
    – saz
    Sep 10, 2019 at 9:19

1 Answer 1

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Hint :

Let $u \in \mathcal L^1(P)$. Then, $\int |u|dP = \int (\int_{0}^{|u|} 1 dt)dP$. Switch the integrals (why can you do this?) and simplify to get $\int_0^\infty P(|u| \geq t) dt = \int |u|dP$. Conversely, note that if this integral involving $t$ is finite, then $u \in \mathcal L^1(P)$.

Now, use a standard integral-dominates-sum-dominates-integral type argument to argue that $\int_0^\infty P(|u| \geq t)dt$ and $\sum_{j \geq 0} P(|u| \geq j)$ either shoot to infinity together or are both finite.

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