Just have been trying to approach this problem from Resnick's book on probability but have got no clue so far.
The problem is like this:
We are giving two random variables X, Y on the same space $(\Omega, \mathcal{B})$, and we are asked to show: $\sup_{A \in \mathcal{B} } | P[X\in A] - P[Y\in A] | \leq P[X \neq Y] $.
What I have thought:
My intuition is that maybe X and Y have the same distribution, although I don't see how the distribution of the two RVs plays a role here.
For the RHS I can say that if we set $P[X \neq Y] = \epsilon$, and we can check that the LHS $< \epsilon$ we may be able to get this done.
I realized that, if X, Y are random variables, then both satisfy the mapping:
$ X:(\Omega, \mathcal{B}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})) $
Then, $A \in \mathcal{B}(\mathbb{R})$, so I'm confused why the problem states that $A \in \mathcal{B}$.
Any solid hint please?
It suffices to prove that for any $A\in \mathcal B$, $$| P[X\in A] - P[Y\in A] | \leq P[X \neq Y]$$
By symmetry, it suffices to prove that $$P[X\in A] - P[Y\in A] \leq P[X \neq Y]$$
which rewrites as $P[X\in A] \leq P[Y\in A] + P[X \neq Y]$.
The last inequality follows from $$\begin{align} P[X\in A] &= P[X\in A \;\cap\; X \neq Y] + P[X\in A \;\cap\; X = Y]\\ &= P[X\in A \;\cap\; X \neq Y] + P[Y\in A \;\cap\; X = Y]\\ &\leq P[ X \neq Y] + P[Y\in A ] \end{align}$$
and we're done.