Let $X$ and $Y$ be two i.i.d random variables. I am trying to prove that $\mathbb{P}(X<Y)=\mathbb{P}(Y<X)$. The author claims that this is true by symmetry. But I am trying to prove this in a rigorous way.
I haven't studied joint distributions yet. Here is my attempt for the proof.
Let $\Omega$ be the sample space of interest. Then we have the following.
$\mathbb{P}(X<Y)=\mathbb{P}\left(\{\omega\in\Omega:X(\omega)<Y(\omega)\}\right)$
$\mathbb{P}(Y<X)=\mathbb{P}\left(\{\omega\in\Omega:Y(\omega)<X(\omega)\}\right)$
One way to show that $\mathbb{P}(X<Y)=\mathbb{P}(Y<X)$ is to show that the following equation holds:
$$\begin{equation*}\mathbb{P}\left(\{\omega\in\Omega:X(\omega)<Y(\omega)\}\right)=\mathbb{P}\left(\{\omega\in\Omega:Y(\omega)<X(\omega)\}\right)\end{equation*}\cdots\cdots(1)$$
However, I couldn't continue further using the above approach. Instead, I am using the following method, but it works only if $X$ and $Y$ are discrete random variables.
Let $A:=\{x\in\mathbb{R}:X(x)>0\}$. Then $A$ is the support of $X$, and also $Y$ since $X$, $Y$ are i.i.d.
Let $B:=\{(x,y)\in A\times A:x<y\}$.
We then have
$\begin{equation*}\begin{split}\mathbb{P}(X<Y)&=\sum_{(x,y)\in B}\mathbb{P}(X=x,Y=y)\\&=\sum_{(x,y)\in B}\mathbb{P}(X=x)\mathbb{P}(Y=y)\quad\quad\quad \left(X,Y \text{ are independent}\right)\\ &=\sum_{(x,y)\in B}\mathbb{P}(Y=x)\mathbb{P}(X=y)\quad\quad\quad \left(X,Y \text{ are identically distrbuted}\right)\\&=\mathbb{P}(Y<X)\end{split}\end{equation*}$
The above proof holds when $X$ and $Y$ are discrete. But how do I approach this problem when $X$ and $Y$ are continuous random variables? Also, is it possible prove the above fact by directly showing that the equation $(1)$ holds?
$P(X<Y) =\int_{-\infty}^{\infty}\int_{-\infty}^y f(x) f(y) \;dx \;dy$
Swap the order of integration
$\int_{-\infty}^{\infty}\int_{-\infty}^x f(x) f(y) \;dy \;dx = P(Y<X)$