Showing $\big|E[X]-E[Y]\big|\leq 2\Delta$

Question

Am studying the following proposition:

Let $X,Y$ be real random variables on the probability space $(\Omega,\mathcal F,P)$ satisfying $\big|X\big|\leq 1$, $\big|Y\big|\leq 1$, and let $\Delta=\sup_{B\in\mathcal{B}(\mathbb R)}\big|P(X\in B)-P(Y\in B)\big|$. Then

$$\bigg|E[X]-E[Y]\bigg|\leq 2\Delta$$

Proof.

Let $\nu(B)=P(X\in B)-P(Y\in B)$ for all $B\in\mathcal{B}(\mathbb R)$. Then $\nu$ is a finite signed measure on $(\mathbb R,\mathcal{B}(\mathbb R))$. We have

$$\bigg|E[X]-E[Y]\bigg|=\bigg|\int_{[-1,1]} t \, dP_X-\int_{[-1,1]} t \,dP_Y\bigg|=\bigg|\int_{[-1,1]} t \, d\nu \bigg|$$

$$\leq \int_{[-1,1]} |t| \, d|\nu|\leq \int_{[-1,1]} 1 \, d|\nu|=|\nu| ([-1,1])$$

Let $\nu=\nu^+-\nu^-$ be the Jordan decomposition of $\nu$. Since $\nu([-1,1])=0$, we have $|\nu| ([-1,1])=2 \nu^+([-1,1])$. Finally we have

$$\nu^+([-1,1])=\sup_{ B\in\mathcal{B}(\mathbb R),B\subset[-1,1]}\nu(B)=\sup_{ B\in\mathcal{B}(\mathbb R)}\nu(B)\leq \sup_{ B\in\mathcal{B}(\mathbb R)}|\nu(B)|=\Delta$$

Question: It seems that the absolute values in $\Delta$ are not necessary. Is this correct?

Answer 1

$P(X \in B)-P(Y\in B)=[1-P(X \in B^{c}] -[1-P(Y \in B^{c}]=P(Y \in B^{c})-P(X\in B^{c})$. When $B$ vaires over $\mathcal B(\mathbb R)$ so does $B^{c}$. So when you take the supremum the absolute value sign becomes unnecessary.