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Recall that a standard Borel space is a measurable space $(X,\mathcal{M})$ (i.e. a set with a $\sigma$-algebra) such that there exists a 1-1 bimeasurable map $\phi$ from $(X,\mathcal{M})$ to $[0,1]$ (the latter equipped with its Borel $\sigma$-algebra $\mathcal{B}_{[0,1]}$). It is known that any Borel subset of a complete separable metric space is a standard Borel space.

Suppose now that $E$ is a non-Borel subset of $[0,1]$ (or any other complete separable metric space), such as the Vitali set. We can equip $E$ with the $\sigma$-algebra $\mathcal{M}$ induced by its inclusion into $[0,1]$, namely $$\mathcal{M} = \{ B \cap E : B \in \mathcal{B}_{[0,1]}\}.$$ $\mathcal{M}$ is also the Borel $\sigma$-algebra generated by the subspace topology on $E$, which is also the metric topology on $E$.

Is it possible that $(E,\mathcal{M})$ is a standard Borel space? I would think not. Clearly the inclusion map $E \hookrightarrow [0,1]$ is not bimeasurable (though it is measurable), but it's less clear that no other injection could be bimeasurable.

Wikipedia gives an example using a set $E$ of outer measure 1 and inner measure 0, and shows indirectly that it cannot be standard Borel by equipping it with the probability measure $P(B \cap E)=m(B)$, noting that the inclusion map $X : E \to [0,1]$ is a uniformly distributed random variable, and observing that $X$ does not admit a regular conditional distribution given itself. However, it is not so clear how to extend this to other non-Borel sets (particularly those with outer measure 0), and in any case it would be nice to have a direct proof if possible.

Thanks!

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up vote 4 down vote accepted

Strictly speaking, a standard Borel space can be finite or countable. Keeping in mind this minor point, A subset of $[0,1]$ (endowed with the restriction of the Borel $\sigma$-algebra) is a standard Borel space if and only if it is a Borel subset of $[0,1]$.

This comes from the fact that the image of a Borel subset by an injective Borel map between two standard Borel spaces is a Borel set. This result is proven for example in Dudley "Real analysis and Probability" or Cohn "measure theory".

Now for the proof. Let $\phi$ be the isomorphism between $([0,1],{\cal B})$ and $(E,{\cal M})$. The space E is a subset of $[0,1]$, so we get a Borel injection from $[0,1]$ to $[0,1]$ whose image is precisely $E$. Hence $E$ is a Borel subset of $[0,1]$.

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Thanks! The result you mention would certainly do it. I looked in Dudley but did not see it; would you mind giving a theorem number? Cohn is checked out of my library but I can also look there eventually. –  Nate Eldredge Nov 8 '10 at 18:50
    
Indeed, this is not stated in Dudley, sorry for pointing in the wrong direction. This is done in Cohn, and also in Kechris "classical descriptive set theory", thm 15.2 p89, online on google books, books.google.com/… Hope that helps –  user6129 Nov 9 '10 at 21:15
    
Thanks very much! –  Nate Eldredge Nov 10 '10 at 17:13
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