Three doubts in the proof of Hölder's inequality

Question

I quote Jacod-Protter (2004).

Theorem (Hölder's inequality) Let $X$, $Y$ be random variables with $\mathbb{E}\{|X|^p\}<\infty$, $\mathbb{E}\{|Y|^q\}<\infty$, where $p>1$ and $\frac{1}{p}+\frac{1}{q}=1$. Then $$|\mathbb{E}\{XY\}|\le\mathbb{E}\{|XY|\}\le\mathbb{E}\{|X|^p\}^{\frac{1}{p}}\mathbb{E}\{|Y|^q\}^{\frac{1}{q}}$$ Proof Without loss of generality, we can assume $X\ge0, Y\ge0$ and $\mathbb{E}\{X^p\}>0$, since $\mathbb{E}\{X^p\}=0$ implies $X^p=0$ a.s., and thus $X=0$ a.s. and there is nothing to prove. [...]

---

Three questions about initial statement of the proof:

1. Why are we allowed to say that "Without loss of generality, we can assume $X\ge0, Y\ge0$"? Why does not such an assumption have an impact on the proof of the theorem?
2. Why "$\mathbb{E}\{X^p\}=0$ implies $X^p=0$ a.s."? Couldn't $X^p$ be distributed in a way such that its expectation is $0$, but it is not almost surely equal to $0$ as a random variable?
3. Finally, why $X^p=0$ a.s. implies $X=0$ a.s.? Is this a consequence of the fact that $p>1$, hence it cannot be possible that $X^p=X^{-\infty}=0$, hence meaning that in this case $$X^p=0\text{ a.s.}\iff X=0\text{ a.s.?}$$

Answer 1

1. Well, the first inequality is trival and the second inequality only deals with $|XY|=|X|\cdot|Y|$, $|X |$ and $|Y|$, so it's really a statement about $|X|$ and $|Y|$ which are positive random variables.

2. No, this cannot be, since $Z:=X^p$ is assumed positive. Indeed, $P(Z\neq 0)=\lim_{n\to\infty} P(Z\geq \frac{1}{n})$ by continuity of measures and you have that $0=\mathbb{E}Z\geq n P(Z\geq 1/n)$ by Markov's Inequality, so $P(Z\geq \frac{1}{n})=0$ and we get that $P(Z\neq 0)=0$.

3. This is just the fact that $\{X\neq 0\}=\{X^p\neq 0\}$.