$X,Y$ has same distribution function $\implies P(X=Y)=1$ ? Suppose $X,Y$ are two random variables with same distribution function. Denote $F_X$ for the distribution function of $X$ and $F_Y$ the distribution function of $Y$, then $F_X(t)=F_Y(t)$ for all $t\in\mathbb{R}$. I want to know if $P(X=Y)=1$. If this is not the case, could you show me some counter example. 
PS: if the statement is valid, I'm looking for some argument not using expectations. 
Thank you.
 A: The statement is not valid. 
Flip a fair coin.
If $X$ denotes the number of heads and $Y$ the number of tails then they have the same distribution, but $X\neq Y$ hence $P(X=Y)=0$.
A: This doesn't hold at all if $X$ and $Y$ are non-degenerate. Consider for example the space $\Omega = \{0, 1\}$ with the uniform distribution and the random variables $X(\omega) = \omega$ and $Y(\omega) = 1 - \omega$. They both have the distribution, but $P(X = Y) = 0$.
The reason for this is that the connection between the distribution of a random variable and the probability measure on the domain of $X$ and $Y$ is very weak. For example, two random variables can be defined on completely different probability spaces, but still have the same distribution.
A: Let $X$ be normally distributed and $Y\equiv -X$. Clearly, $X$ and $Y$ have the same distribution, because the normal distribution function is symmetric about $0$: 
\begin{align*}
F_Y(t)=&\,\mathbb P(Y\leq t)=\mathbb P(-X\leq t)=\mathbb P(X\geq -t)=1-\mathbb P(X<-t)=1-\mathbb P(X\leq -t)\\=&\,1-F_X(-t)=F_X(t).
\end{align*}
However, $$\mathbb P(X=Y)=\mathbb P(2X=0)=\mathbb P(X=0)=0.$$
