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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.

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If you require $\mu$ to be $\sigma$-additive and $X$ is countable then the measure is either atomic or zero. – Asaf Karagila May 17 '12 at 8:34
@AsafKaragila: If $X$ is countable, then the fact that $\mu$ is continuous implies that it is zero. But I'm still confused. – ragrigg May 17 '12 at 8:37
@ggirgar: There are no cntinuous probability measures on a countable set, so the result holds vacuously for such spaces. – Michael Greinecker♦ May 17 '12 at 8:42
@MichaelGreinecker: OMG, I apologize. I knew that it was something really simple. We must have $\mu(X) = 1$... sorry. – ragrigg May 17 '12 at 8:54

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