Artificial intelligent assistant

A doubt in Probability essentials by Protter ! Please see part \(v\) Why is the first inequality valid(In part (v))? I mean I know that we can show that if $X \in L^2 \implies X \in L^1 $ by using **Holder's Inequality**. This inequality does not make any sense to me howsoever. I would be grateful if someone could explain me why! Thanks

It seems, at least to me, that taking the absolute values off there after the first inequality is not necessary, or maybe not even desirable. Note that $\Omega=\\{\omega:|X(\omega)|<1\\}\cup\\{\omega:|X(\omega)|\geq 1\\}$ and these sets are disjoint, so the sum is only divided to two different parts there. In more detail:

1. In the first inequality he divides the sum over those two disjoint sets. You can keep absolute values here on both sums on the right and we have an equality instead of an inequality.

2. Then he uses the estimate on $|X(\omega)|<1$ in the first sum of the second inequality by replacing $|X(\omega)|p_{\omega}$ with $p_{\omega}$, and in the second sum as $|X(\omega)|\geq 1$ then $|X(\omega)|\leq |X(\omega)|^{2}$.

3. He omitted this step but in the second inequality he finally replaces the sum over a subset of $\Omega$ by sum over the whole $\Omega$ in both sums.

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