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Suppose that two random variables $X,Y$ are symmetric, in that:

$$ P(X \leq x, Y \leq y) = P(X \leq y, Y \leq x) $$

I read that if this is satisfied, then the marginal distributions of $X$ and $Y$ are the same. Can anyone see why this is the case? Thanks.

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    Take the limit $x \to +\infty$. – 2017-02-16
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    Why is it that $\lim_{x \to \infty}P(X \leq x, Y \leq y) = P(Y \leq y)$? Is there a standard result in probability for this? – 2017-02-16
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    Because it is $\mathsf P(X\in \Bbb R, Y\in (-\infty;y])$, the probability that $Y$ is at most $y$ and $X$ is any real number. @user321627 – 2017-02-16

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Your supposition can also be written $$ P \left( X=x, Y=y \right) = P \left( X=y, Y=x \right). $$ The marginal distribution is $$ P \left( X=x\right) = \sum_{y} P \left( X=x, Y=y \right) = \sum_{y} P \left( X=y, Y=x \right) = P \left( Y =x\right)$$ i.e. the marginal distributions are the same.