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 $F$.
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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 $F$.
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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.
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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.
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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?
Best Answer
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.