We have the following theorem:
Theorem Let $X$ be a standard Borel space and $\mu$ be a continuous probability Borel measure. Then there is a Borel isomorphism $f: X \to [0,1]$ such that $f\mu$ is the Lebesgue measure on $[0,1]$.
By the Isomorphism Theorem for standard Borel spaces, a typical proof of this theorem begins saying that we can assume $X$ to be $[0,1]$. My question is the following: why is it true when $X$ is countable? In that case we cannot even have a bijection between $X$ and $[0,1]$, and consequently there cannot be such a Borel isomorphism.