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

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.

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

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.