Prove that $\mathbb P(X=Y)=0$

Question

Prove that $\mathbb P(X=Y)=0$ if we know that $X,Y$ are independent random variables and for every $a$ we have $\mathbb P(X=a)=0$.

I did this task but I think that my proof is too easy and I afraid that it isn't correct answer: $$\mathbb P(X=Y)=\sum_{a\in \mathbb R} \mathbb P(X=a \cap Y=a)=\sum_{a\in \mathbb R} \mathbb P(X=a)\cdot\mathbb P(Y=a)=\sum_{a\in \mathbb R} 0\cdot \mathbb P(Y=a)=0$$

Can anyone check it?

Answer 1

A few more details on another answer (Fubini's Theorem not used): \begin{eqnarray*} {\bf P}(X=Y) &=& \int_{\mathbb{R}}{\bf P}(X=Y|Y=y)dF_Y(y)\qquad \text{(by Law of Total Probability)}\\ &=& \int_{\mathbb{R}}{\bf P}(X=y|Y=y)dF_Y(y)\qquad \text{(because of conditioning on }Y\text{)}\\ &=& \int_{\mathbb{R}}{\bf P}(X=y)dF_Y(y)\qquad \text{(because of independence of } X,Y \text{)}\\ &=& \int_{\mathbb{R}}0 dF_Y(y)\qquad \text{(because of assumption on } X\text{)}\\ &=& 0 \end{eqnarray*}

Answer 2

Your argument doesn't work because you are using 'uncountable additivity'.

$P(X=Y)=\int P(X=a) dF_Y(a)=\int 0dF_Y(a)=0$. The first equality is a consequence of Fubini's Theorem.