# 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 $[0,1]$.