I have a question about the notation used in this problem please dont solve the problem for me.

Show that $||XY|{|}_{1}=||X|{|}_{1}||Y|{|}_{1}$ for independent r.vs $X$ and $Y$. Show further that if $X$ and $Y$ are also integrable, then $E(XY)=E(X)E(Y)$

I don't understand what the difference between $||X|{|}_{1}$ and $E(|X|)$. I assume we're talking about the ${L}^{1}(\mathbb{P})$ norm? But then

$||X|{|}_{1}={\int}_{\mathrm{\Omega}}|X(\omega )|d\mathbb{P}(\omega )=E(|X|)$

So if this is finite, then $E(X)$ exists therefore it is integrable already (so why "also"?) Therefore the first "show that" would follow from the second

Show that $||XY|{|}_{1}=||X|{|}_{1}||Y|{|}_{1}$ for independent r.vs $X$ and $Y$. Show further that if $X$ and $Y$ are also integrable, then $E(XY)=E(X)E(Y)$

I don't understand what the difference between $||X|{|}_{1}$ and $E(|X|)$. I assume we're talking about the ${L}^{1}(\mathbb{P})$ norm? But then

$||X|{|}_{1}={\int}_{\mathrm{\Omega}}|X(\omega )|d\mathbb{P}(\omega )=E(|X|)$

So if this is finite, then $E(X)$ exists therefore it is integrable already (so why "also"?) Therefore the first "show that" would follow from the second