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I've tried figuring this out and searching the net on this for 5 hours, but I can't get it. Every source says it's trivial, but I must be missing something because I have pages of work that don't lead me anywhere.

How do you prove $||X+Y||_\infty\leq||X||_\infty+||Y||_\infty$. The definition of the $\infty$-norm is $||X||_\infty=\sup\left\{\alpha|P(|X|>\alpha)>0 \right\}$. However, I can't figure out how to combine the $X$ and $Y$ norm in terms of sets with measure greater than zero.

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Is it the triangle inequality or the measure part causing the problem? –  user13838 Oct 13 '11 at 23:53
    
Maybe it's easier if you write $\|X\|_\infty = \inf{\{\alpha\mid P(|X| \gt \alpha) = 0\}}$? –  t.b. Oct 14 '11 at 0:00

2 Answers 2

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Define the notation $A(X) = \{ \alpha\colon P(|X|>\alpha)>0 \}$, so that $\|X\|_\infty = \sup A(X)$; we want to show that $\sup A(X+Y) \le \sup A(X) + \sup A(Y)$. Usually when trying to bound a sup from above, it's a good idea to name a generic element $\beta \in A(X+Y)$ and show that $\beta \le \sup A(X) + \sup A(Y)$. It suffices to show that there exists $c\in A(X)$ and $d\in A(Y)$ such that $\beta\le c+d$. (In some situations, it helps to show that for every $\epsilon>0$, one can find $c\in A(X)$ and $d\in A(Y)$ such that $\beta<c+d+\epsilon$.)

So let $\beta\in A(X+Y)$, so that $P(|X+Y|>\beta)>0$. By the triangle inequality this implies that $P(|X|+|Y|>\beta)>0$. But notice that $|X|+|Y|>\beta$ if and only if there exist real numbers $c$ and $d$ such that $c+d>\beta$ and $|X|>c$ and $|Y|>d$; in fact, one can restrict $c$ and $d$ to be rational numbers. In other words, the event that $|X|+|Y|>\beta$ is contained in the union of the countable set of events $\big( |X|>c \text{ and } |Y|>d \big)\colon c,d\in\mathbb Q$. Since this countable union has positive probability, one of the individual events $\big( |X|>c \text{ and } |Y|>d \big)$ must itself have positive probability; in particular, each of $|X|>c$ and $|Y|>d$ has positive probability, so that $c\in A(X)$ and $d\in A(Y)$. Since $\beta<c+d$, the proof is complete.

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To prove $$\sup\left\{\alpha\mid P(|X+Y|>\alpha)>0 \right\}\leq||X||_\infty+||Y||_\infty$$ it suffices to prove for every $\alpha$ that $$P(|X+Y|>\alpha)>0 \Longrightarrow ||X||_\infty+||Y||_\infty \ge\alpha$$ and for that we only need to prove for arbitrary $\epsilon>0$ that $$P(|X+Y|>\alpha)>0 \Longrightarrow ||X||_\infty+||Y||_\infty\ge\alpha-4\epsilon$$ Thus, assume the left-hand side, and consider the conditional probablity distribution of $(|X|,|Y|)\in\mathbb R^2$ given $|X+Y|>\alpha$. Since $\mathbb R^2$ is covered by countably many balls of radius $\epsilon$, there must be one such ball that contains positive probability mass, so there exist $x,y$ such that $$P(|X|>x-\epsilon)>0 \land P(|Y|>y-\epsilon)>0$$ and hence $||X||_\infty+||Y||_\infty\ge x+y-2\epsilon$.

Furthermore, we must have $x+y\ge\alpha-2\epsilon$, because otherwise the triangle inequality prevents the ball from containing any $(|X|,|Y|)$ with $|X+Y|>\alpha$. Therefore $$||X||_\infty+||Y||_\infty\ge\alpha-4\epsilon$$ as promised.

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