Prove $P(X=Y) = 0$ when $X$ and $Y$ are two random variables on the same probability space. Suppose that $X$ and $Y$ are two independent random variable on the same probability space. We assume: 
$$P(X=x) = P(Y=x) = 0$$
and I want to prove $P(X=Y) = 0$. 
But we don't know whether the distribution of r.v-s are continuous or not. 
I tried to solve with the fact that $$
P(X = Y) = E(1_{X=Y})$$ 
But unfortunately it didn't work well. How to solve this problem in measure theory-wise?
 A: Let $\mu$ be the distribution of $X$ and $\nu$ the distribution of $Y$.  "Independent" tells us that
$$
\mathbb P\big[(X,Y) \in E\big] = (\mu \otimes \nu) (E) 
$$
for all Borel sets $E \subseteq \mathbb R \times \mathbb R$.  In particular, when $E$ is the "diagonal" $E = \{(x,x) : x \in \mathbb R\}$ we get
$$
\mathbb P\big[X=Y\big] =\mathbb P\big[(X,Y) \in E\big] = (\mu \otimes \nu) (E) 
$$
And by Fubini (or Tonelli),
$$
(\mu \otimes \nu) (E) = \int_\Omega \nu\{y : (x,y) \in E\}\;d\mu(x)
=\int_\Omega \nu\{x\}\;d\mu(x) =\int_\Omega 0\;d\mu(x) = 0 .
$$
A: Your initial approach is correct and works well:$$\mathbb E1_{X=Y}=\int\int1_{x=y}dF_Y(y)dF_X(x)=\int P(Y=x)dF_X(x)=\int0dF_X(x)=0$$
Also we can switch the order of integration and that makes clear that it is enough already to demand that at least one (not necessarily both) of the independent rv's has continuous distribution.
A: Let $\mu_Y$ the measure induced by $Y$ over $\mathbb R$ (i.e. $\mu_Y(A)=\mathbb P\{Y\in A\}$). Then
\begin{align*}
\mathbb P\{X=Y\}&=\int_{\mathbb R}\mathbb P\{X=y\mid Y=y\}\mu_Y(\mathrm dy)\\
&\underset{(*)}{=}\int_{\mathbb R}\mathbb P\{X=y\}\mu_Y(\mathrm dy)\\
&=0,
\end{align*}
where $(*)$ follow from the independence of $X$ and $Y$.
