Conditional probabilities problem I need help on the last step of a problem I am trying to solve:
$X, Y$ are integrable r.v. and I proved that $$\mathbb{P}((X-Y)(1_{X \geq c}-1_{Y \geq c})=0)=1$$ for any $c\in \mathbb{R}$
I need to prove that $X=Y$ a.s.
I was thinking of doing it by using the law of total probability and conditioning on the two events $\{X=Y\}$ and $\{X\not = Y\}$ 
So that $$1=\mathbb{P}((X-Y)(1_{X \geq c}-1_{Y \geq c})=0)=\mathbb{P}(X=Y) + \mathbb{P}(X\not = Y)\mathbb{P}((1_{X \geq c}-1_{Y \geq c})=0 \ | X\not = Y ) $$
Which then should lead to $\mathbb{P}(X\not = Y)=0$ by arbitrarieness of $c$
Am I completely out of track?
Thanks
 A: Suppose $\mathbb{P}(X \neq Y) \neq 0$.
I am given that, $\forall c \in \mathbb{R}$,
$$
\mathbb{P}(X = Y) + \mathbb{P}(X \neq Y) \mathbb{P}(1_{X \geqslant c} - 1_{Y \geqslant c} = 0 | X \neq Y) = 1
$$
Which suggests that
\begin{align}
\mathbb{P} (X \neq Y) &= \mathbb{P}(X \neq Y) \mathbb{P}(1_{X \geqslant c} - 1_{Y \geqslant c} = 0 | X \neq Y)\\
1 &=\mathbb{P}(1_{X \geqslant c} - 1_{Y \geqslant c} = 0 | X \neq Y)
\end{align}
This holds for all $c \in \mathbb{R}$. Let $c = \frac{X+Y}{2}$, so regardless of the values realised by $X$ and $Y$, either $X < c < Y$ or $Y < c < X$.
So $1_{X \geqslant c} - 1_{Y \geqslant c} \neq 0$ a.s.
This leads to a contradiction, so it must be the case that $\mathbb{P}(X \neq Y) = 0$, and $X = Y$ a.s.
A: Building off ymbirtt answer,  take the intersection of countably many events so that given that $X\neq Y$, almost surely we have that for all rational $c$ either both $X,Y$ are less than $c$ or both greater than $c$. Such an event is impossible by density of rationals if $X$ and $Y$ are different, so it is a contradiction and the the event that $X\neq Y$ happens with probability zero.
