# If a metric space $F$ is Borel isomorphic to a Lusin measurable space, then $F$ must be homeomorphic to a Borel subset of some compact metric space

I'm reading about Lusin and Suslin spaces from page 46 of Dellacherie/Meyer's Probabilities and Potential. Here I restrict myself to metric spaces.

A Borel isomorphism between two measurable spaces is a measurable bijection whose inverse is also measurable. Let $$F$$ be a metric space together with its Borel $$\sigma$$-algebra $$\mathcal F$$. We denote by $$F_\sigma$$ (resp. $$F_\delta$$) the closure of $$F$$ under countable union (resp. intersection). Let $$F_{\sigma \delta} := (F_\sigma)_\delta$$. We denote by $$\mathcal K (F)$$ the collection of all compact subsets of $$F$$.

• Def 1 A subset $$A$$ of $$F$$ is said to be analytic if there exists a compact metric space $$E$$ and $$B \in (\mathcal K (E) \times \mathcal F)_{\sigma \delta}$$, such that $$A$$ is the projection of $$B$$ onto $$F$$.

• Def 2 $$F$$ is said to be Lusin (resp. Suslin, cosuslin) if it is homeomorphic to a Borel subset (resp. an analytic subset, the complement of an analytic subset) of a compact metric space.

Let $$(X, \mathcal X)$$ be a measurable space.

• Def 3 $$X$$ is said to be Lusin (resp. Suslin, cosuslin) if it is Borel isomorphic to a measurable space $$(H, \mathcal H)$$ where $$H$$ is a Lusin (resp. Suslin, cosuslin) metric space.

• Def 4 A subset $$Y$$ of $$X$$ is said to be Lusin in (resp. Suslin, cosuslin) if the measurable space $$(Y, \mathcal{X}|_\mathcal Y)$$ is Lusin (resp. Suslin, cosuslin).

My understanding: Let $$F$$ be a metric space and $$\mathcal F$$ its Borel $$\sigma$$-algebra. Suppose that $$(F, \mathcal F)$$ is Borel isomorphic to a Lusin measurable space $$(H, \mathcal H)$$. Then $$F$$ is Lusin by Def 3. To be logically consistent with Def 2, $$F$$ must be homeomorphic to a Borel subset of some compact metric space.

Could you confirm that if a metric space $$F$$ is Borel isomorphic to a Lusin measurable space then $$F$$ must be homeomorphic to a Borel subset of some compact metric space?

Yes. If a metric space is Borel isomorphic to a Borel subset of a compact metric space, then it must be separable. This follows from this answer together with the fact that every uncountable metrizable Lusin space is isomorphic to $$[0,1]$$, which Dellacherie & Meyer show on page 159 in an appendix. So if $$X$$ is a metrizable Lusin space, it must homeomorphically embed as a subspace of the Hilbert cube $$H=[0,1]^\infty$$, a compact metric space. The canonical injection $$j:X\to H$$ is of course measurable. Now, the image of a measurable injection from a Polish space into a Polish space is a Borel set, so $$j(X)$$ must be a Borel subset of $$H$$ and $$j$$ is a homeomorphism of $$X$$ onto a Borel subset of a metric space.
• No, I very much meant a metric space endowed with its Borel $\sigma$-algebra. Commented Oct 30, 2022 at 21:19
• Thank you so much! I got it. It becomes clear because a measurable space $(X, \mathcal X)$ is Borel isomorphic to some compact metric space IFF $(X, \mathcal X)$ is Borel isomorphic to some Borel subset of some compact metric space IFF $(X, \mathcal X)$ is Borel isomorphic to some Borel subset of some Polish space. Commented Oct 30, 2022 at 23:32