# Can a non-Borel set be a standard Borel space?

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!

Strictly speaking, a standard Borel space can also 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 $$[0,1]$$.
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 $$[0,1]$$.