Is there some notion of "nice" measurable spaces and "nice" maps between them which satisfies the following properties?

- The real line equipped with the Lebesgue $\sigma$-algebra is nice.
- Any translation $\mathbb R \to \mathbb R$ is nice. (Ideally, any continuous map $\mathbb R \to \mathbb R$ is nice, but right now I'm not picky.)
- Any finite or countable set equipped with the discrete $\sigma$-algebra is nice.
- Nice measurable spaces and nice maps form an elementary topos.

Observations:

- If such a topos exists, it is Boolean but does not satisfy the axiom of choice. To prove the latter: Construct the unit circle as a quotient of $\mathbb R$, and then consider the quotient of the unit circle by an irrational rotation. This quotient map cannot split, because if it did the image of its section would be a Vitali set.
- If such a topos does not exist, any proof of its nonexistence must rely on the axiom of choice, since in the Solovay model the category of discrete measurable spaces satisfies the conditions.

Edited to add: I would also be interested in the answer to the same question with the Lebesgue $\sigma$-algebra replaced by the Borel $\sigma$-algebra.

morphismsthat it seems hard to deduce anything about any such topos. The only function that can equalize two distinct translations of $\mathbb{R}$ is the one with empty domain, so the initial object could only be the empty measurable space. But I'm honestly having trouble seeing that the terminal could only be the one-point space $1$ (e.g., we don't know that there are any nice maps to $1$, except the identity on $1$). How do you see the topos would have to be Boolean? $\endgroup$ – Todd Trimble♦ Feb 3 '16 at 5:54