Lebesgue measure of a 1D set over 2D space

Question

I'm trying to prove mathematically that if two continuous real independent random variables $X$ and $Y$ have different densities then they are different.

This is what I did :

$P(X(w)=Y(w))$=$P((X,Y)$$\in A)$ with $A=${$(X(w),Y(w))/X(w)=Y(w)$}

Now I calculate $P((X,Y)\in A)$$=\int_{R^2}$$ _A$$d\lambda_{R^2}$ = $\int_{A}$$d\lambda_{R^2}$$=0$

is this right? since the dimension of A is 1 and the the lebesgue measure is in $R^2$

Answer 1

That is too vague an argument.

$P(X=Y)=\int_{\mathbb R^{2}} 1_A d\mu$ where $A=\{(x,x):x \in \mathbb R\}$ and $\mu (E)=P((X,Y) \in E)$ for all Borel sets $E$ in $\mathbb R^{2}$. By Fubini's Theorem $P(X=Y)=\int \int 1_{\{x\}}(y) f_Y(y)dy f_X(x)dx =\int 0 f_X(x)dx =0$.

I have used the fact that for fixed $x$, $(x,y) \in A$ iff $y=x$.

Incidentally, the result is true even if $X$ and $Y$ do not densities provided $P(X=x)=0$ for all $x$ and $P(Y=y)=0$ for all $y$.