$P[X=Y]=0$ if $X,Y$ are i.i.d. with continuous c.d.f.

Question

I am having lots of trouble proving the following statement:

Let $X,Y$ be two real valued random variables on a probability space $(\Omega,\mathcal{F},P)$. These two variables are independent and identically distributed, iid, in the sense that they share the same distribution $P_X=P_Y$ and the cumulative distribution function $F_X(x)=P_X[X\leq x]$ is continuous. The probability measure $P$ might not necessarily be absolutely continuous wrt a sigma-finite measure. Therefore there might not be a Radon-Nikodym derivative with respect to a sigma-finite measure. If there was a Radon-Nikodym derivative with respect to a sigma-finite measure, it is easier to show what I want.

I need to prove that, under the above conditions $P[X=Y]=0$

Any insights?

Best regards,

Juan Manuel

Answer 1

For any increasing sequence $-\infty=x_0<x_1<\ldots < x_n=\infty$, we have $$\tag0P[X=Y]\le \sum_{i=1}^n P[x_{i-1}<X\le x_i,x_{i-1}<Y\le x_i]=\sum_{i=1}^n (F_X(x_i)-F_X(x_{i-1}))^2.$$ Since $F_X\colon\mathbb R\to[0,1]$ is continuous and at least covers $(0,1)$, we can pick $x_i$ so that $F_X(x_i)=\frac in$. Then $(0)$ gives us $P[X=Y]\le \frac 1n$.

Answer 2

Here is a big machinery argument. Since the cdf is continuous, the probability of each point is zero. let $D$ be the diagonal on $\mathbb{R}^2$. By Fubini's theorem, you get

$$P(X=Y)=\int_\mathbb{R^2}1_DdP_{(X,Y)}=\int_\mathbb{R}\int_\mathbb{R}1_D(x,y)~dP_X(x)~ dP_Y(y) =\int_\mathbb{R}0~dP_Y(y)=0.$$