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I am reading the book Discrete-Time Markov Control Processes: Basic Optimality Criteria by Onésimo Hernández-Lerma and Jean Bernard Lasserre. On page 14 of the book, they define a Markov control model which is the object studied throughout the book. The state space is denoted by $X$ and and is assumed to be a Borel space throughout the book. They define a Borel space in several different spots (e.g., page xiv, page 13, page 169, etc). I realized that I did not fully understand their definition of a Borel space so I figured I would ask here. Here are the first two paragraphs on page 169 of the book:

A topological space will always be endowed with the Borel $\sigma$-algebra $\mathcal{B}(X)$, that is, the smallest $\sigma$-algebra of subsets of $X$ that contains all of the open sets in $X$. Thus, when referring to either sets of functions, "measurable" means "Borel-measurable."

A Borel subset of a complete and separable metric space is called a Borel space. A Borel subset of a Borel space is itself a Borel space. Examples of Borel spaces are..."

How it is read above, the definition translates to this: "Let $(Y,d)$ be a complete and separable metric space. We call any $X\in\mathcal{B}(Y)$ a Borel space."

There are two issues I have with this definition. First, the use of the word "space" typically means there is some additional structure on it, i.e., a $\sigma$-algebra, topology, metric, etc. But, as stated, there is no requirement of any additional structure. My assumption is that some additional structure (most likely an associated $\sigma$-algebra) is implicit in their definition. Second, they do not go on to explain why $X$ being subset of both a "complete" and "separable" metric space is important.

My guess is that the definition should really be the following: "Let $(Y,d)$ be a complete and separable metric space, $X\in\mathcal{B}(Y)$, and $\mathcal{B}(X):=X\cap \mathcal{B}(Y)$. We call the measurable space $(X, \mathcal{B}(X))$ a Borel space."

This new definition resolves my first concern, but I still do not have good insight into why "complete and separable" is important. After all, a subspace of a complete metric spaces must be closed in order to be complete as well. So there is no guarantee that $X$ is complete in its own right. (Subspaces of separable metric spaces are separable, however.)

Wikipedia says some things about "standard Borel spaces," see this link. But I couldn't really connect this to my question. I also did a literature search on Google Scholar. I found some articles that use the same definition as given in this book but without additional explanation. Hoping to get a more proper definition and explanation so that I can proceed with my reading.

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  • $\begingroup$ The topological structure is extra structure. Complete means it contains all its limit points and separable means given any two points I can find a neighborhood of each point which are disjoint from each other. In practice complete ensures solutions will exist and separable ensure they're unique. The Nested Interval theorem is the inspiration here and you don't need algebraic properties to move forward so they're typically not used till they're required. $\endgroup$ Commented Nov 19, 2022 at 23:13
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    $\begingroup$ @CyclotomicField I am still a bit confused. It says $X$ is a subset of a complete and separable space, not the space itself. So I do not see why $X$ must also be complete if that is what you are getting at. I may be mistaken. Also, why do you think that extra structure is the topology? So you think they are taking a Borel set $X\in\mathcal{B}(Y)$, letting $\tau_X := X\cap\tau_Y$ where $\tau_Y$ is the metric topology on $Y$ induced by $d$, and defining the topological space $(X, \tau_X)$ to be a "Borel space"? $\endgroup$
    – Satana
    Commented Nov 19, 2022 at 23:28

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You said yourself that a topological space is by default endowed with its Borel $\sigma$-algebra, so yes when one says that $X$ is a Borel space it should be understood $(X,\mathcal B(X))$ is a Borel space.

I did not read the book, but since it speaks of Markov processes I assume it uses conditional distribution. So for random variables $X$ and $Y$ valued in a space $E$, they would need to be able to consider the conditional distribution of $Y$ given $X$. But the existence of such a thing requires some assumption on $E$. For example the existence is guaranteed when $E$ is a Borel subset of a Polish space.

For the record, the most general assumption on $E$ I know for the conditional distribution to exist would be that there exists a measurable isomorphism (that is a measurable bijection whose inverse is also measurable) between $E$ and a Borel subset of $\mathbb R$. It happens that any Borel subset of a Polish space satisfies this, so we can restrict ourselves to this kind of spaces. In practice it is enough for most applications.

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  • $\begingroup$ Thank you so much! Am I correct with this statement: A complete and separable metric space is itself a Polish space (trivially because it is already endowed such a metric)? This would then explain it because then the state space would be a subset of a Polish space and these conditional distributions would exist. Do you happen to have a reference for this existence result on the conditional distributions of RVs? $\endgroup$
    – Satana
    Commented Nov 19, 2022 at 23:45
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    $\begingroup$ Yes you're correct. For a reference you have for instance Kallenberg's book: Foundations of Modern Probability. See Theorem 8.5 for the existence of a conditional distribution. Be aware that what he calls "Borel" here is defined page 14, which is at first glance a definition of Borel space different of yours, but actually encompasses it by Theorem 1.8 (a Polish space is Borel, from which you can deduce that a Borel subset of a Polish space is isomorphic to a Borel space and is therefore Borel). $\endgroup$
    – Will
    Commented Nov 20, 2022 at 0:25
  • $\begingroup$ Thank you so much. You are very helpful. $\endgroup$
    – Satana
    Commented Nov 20, 2022 at 0:52
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    $\begingroup$ This answer is helpful, but after reading it I still don't know the definition of "Borel space". Does it mean the same as "standard Borel space"? Or that I just take any measurable subset of a complete and separable metric space, give it the subspace topology, and then form the Borel $\sigma$-algebra? Or are those the same? (Unfortunately Google doesn't seem to be much help with this question.) $\endgroup$
    – N. Virgo
    Commented Jan 21, 2023 at 3:04
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    $\begingroup$ @Satana I just checked theorem 1.8 of Kallenberg's 'Foundations of Modern probability' (3rd ed.) and it seems to prove that any standard Borel space (as you defined them) is measurably isomorphic to a Borel subset of $[0,1]$ (meaning your definition of Borel space, but with $Y=[0,1]$). From this I would guess that the spaces you describe are exactly standard Borel spaces. I think the terminology confusion is probably because some authors use "Borel space" to just mean any measurable space, so I guess the word "standard" got added to disambiguate. $\endgroup$
    – N. Virgo
    Commented Nov 1, 2023 at 8:17

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