Prove random variables are not almost surely equal, but are equal in distributions

Question

Suppose we have a probability space $([0,1],\mathcal{B},m)$ where $m$ is the Lebesgue measure of the unit interval.

With this, how would I prove that random variables, say $X,Y: [0,1]\rightarrow \mathbb{R}$ are not almost surely equal, but are equal in distribution?

From Wiki: Two random variables $X$ and $Y$ are not equal almost surely iff $\mathbb{P}(X\ne Y)>0$, while they are equal in distribution if they have the same distribution functions: $\mathbb{P}(X\le x)=\mathbb{P}(Y\le x)$

Answer 1

If you mean to ask:

"how would I prove that random variables $X,Y$ exist that are not almost surely equal but do have equal distribution"

then let $X,Y:[0,1]\to\mathbb R$ be prescribed by $x\mapsto x$ and $x\mapsto 1-x$ respectively.

Here $\{X=Y\}=\left\{\frac12\right\}$ so that $P(X\neq Y)=1$ and it is evident that $X$ and $Y$ both have uniform distribution on $[0,1]$.