If $X$ and $Y$ are i.i.d. random variables, is $P(X < Y) = P(Y < X)$? I am trying to solve the following probability problem:

If $X$ and $Y$ are independent and identically distributed (i.i.d.) random variables, is $P(X < Y) = P(Y < X)$?

Source: https://projects.iq.harvard.edu/files/stat110/files/strategic_practice_and_homework_5.pdf (Pg 2, Qn 2)
The answer provided on Pg 7 (of the pdf) states:

If $X$ and $Y$ are i.i.d., then $P(X < Y) = P(Y < X)$ by symmetry: we can
  interchange $X$ and $Y$ since both are the probability of one draw from the
  distribution being less than another, independent draw.

However, my intuition tells me that since $X$ and $Y$ could be entirely different functions, the above explanation seems a little "handwavy" (e.g. $Y$ could take on values that are very much larger than $X$). So I guess I need a little algebraic proof to convince myself but I am unable to do so on my own.
Could anyone please advise me on how to prove why this statement is true/false via an algebraic proof?
 A: $P(X<Y)=P(Y-X>0) =P(X-Y >0)=P(X>Y)$ where the second equality follows from the fact that the joint distribution of $(X,Y)$ is same as the joint distribution of $(Y,X)$.
To add more details let $E=\{(x,y): y>x\}$. Then $P(Y-X >0)=P((X,Y) \in E)=P((Y,X) \in E)=P(X-Y)>0$. 
To prove that $(X,Y)$ and $(Y,X)$ have the same distribution note that $P((X,Y) \in A \times B)=P(X\in A)P(Y \in B)=P(X\in B)P(Y \in A)=P((Y,X) \in A \times B).$ In the first step I used independence and in the second step I used the fact that $X$ and $Y$ have the same distribution. 
If $\mu$ and $\nu$ are measures on the Borel sets of $\mathbb  R^{2}$ such that $\mu (A \times B)=\nu (A \times B)$ for all Borel sets $A$ and $B$ in $\mathbb  R$ the $\mu=\nu$. This follows from standard arguments in measure theory: the class of finite disjoint unions of measurable rectangles forms a field which generates the Borel sigma field on $\mathbb  R^{2}$. Hence $(X,Y)$ and $(Y,X)$ have the same distribution. 
A: Let $P_X$ and $P_Y$ denote the pushforward of $P$ by $X$ and $Y$ respectively. Since $X$ and $Y$ have the same distribution, $P_X$ and $P_Y$ are equal to some common $Q$: $P_X=P_Y=Q$. Now, note that
$$\begin{align}
P(X<Y) &= \int_{\mathbb R^2} 1_{x<y} dP_{(X,Y)}(x,y) \tag{1} \\
&= \int_{\mathbb R}\int_{\mathbb R} 1_{x<y} dP_{X}(x) dP_{Y}(y) \tag{2}\\
&= \int_{\mathbb R}\int_{\mathbb R} 1_{x<y} dQ(x) dQ(y) \\
&= \int_{\mathbb R^2} 1_{y<x} dQ(y) dQ(x) \tag{3}\\
&= P(Y<X)
\end{align}$$
$\text{(1)}$: Law of the unconscious statistician
$\text{(2)}$: Independence of $X$ and $Y$
$\text{(3)}$: Renaming the variables of integration
Here is an additional way. Note the equalities between measures
$$\begin{align}
P_{(X,Y)} &= P_X\otimes P_Y \tag 1 \\
&= P_Y\otimes P_X \tag 2 \\
&= P_{(Y,X)}
\end{align}$$
$\text{(1)}$: Independence of $X$ and $Y$
$\text{(2)}$: $X$ and $Y$ have the same distribution, thus $P_X=P_Y$ and $P_Y=P_X$
Consequently, $(X,Y)\stackrel{(d)}=(Y,X)$. Letting $f:(x,y)\mapsto 1_{x<y}$, composing with this measurable function preserves equality in distribution, so that $f(X,Y)\stackrel{(d)}=f(Y,X)$ and taking expectations
$$E[f(X,Y)]=E[f(Y,X)] \iff P(X<Y) = P(Y<X).$$
