I met the following definition of standard Borel spaces in Durrett's probability theory book (slightly rephrased):

$(S,\mathcal{S})$ is said to be standard Borel if it is isomorphic (as a measurable space) to a Borel subset of $\mathbb{R}$ equipped with its Borel sigma algebra.

However, this is different from the other definitions I've seen in other sources: e.g., "being isomorphic to a separable complete metric space with the Borel σ-algebra".

Durrett's definition relies on a special space, namely $\mathbb{R}$. Are these two definitions equivalent? I tried to check a descriptive set theory book, but I cannot find out a definite answer.

Thanks in advance!

  • $\begingroup$ Why do you think so? @LukasHeger (for example, a singleton set is standard Borel by both definitions) $\endgroup$
    – J. Doe
    May 15 at 15:04
  • $\begingroup$ They are equivalent. I find "A classical course in descriptive set theory" by Kechris a good reference. $\endgroup$
    – dialegou
    May 15 at 16:07
  • $\begingroup$ Dear @dialegou, could you point to (a) specific theorem(s) in that book? $\endgroup$
    – J. Doe
    May 16 at 6:03
  • $\begingroup$ Dear @J. Doe Corollary 13.4 and Theorem 15.6 $\endgroup$
    – dialegou
    May 16 at 11:29

1 Answer 1


I think these are equivalent but I agree that books in descriptive set theory do not make these things clear.

From the Srivastava book A course on Borel sets we have:

"A standard Borel space is a measurable space isomorphic to a Borel subset of a Polish space."

Theorem 3.3.13 "Any two uncountable standard Borel spaces are Borel isomorphic."

Cor 3.3.16 "Two standard Borel spaces are Borel isomorphic if and only if they are of the same cardinality."

For some reason, the book does not explicitly state the equivalence of Durrett's definition. While the loaded terminology always leaves some shadow of doubt, I believe the theorem and its corollary imply the equivalence you seek.

Areas of uncertainty:

Is Srivastava's definition of "standard Borel space" equivalent to your "isomorphic to a separable complete metric space with the Borel σ-algebra" definition? Probably, but there is room for doubt. From wikipedia we find that "a Polish space is a separable completely metrizable topological space." Now we wonder about the relation between "metrizable" and "metric space" and if "completely metrizable" is the same as "complete."

We also observe that one definition talks about subsets, the other does not. The set of rationals $\mathbb{Q}$ is a Borel subset of $\mathbb{R}$. But it is not complete, I do not think, unless the term "complete" itself has shades of meaning in different areas of mathematics. Of course we could always map $\mathbb{Q}$ to $\mathbb{N}$ (and $\mathbb{N}$ is complete).

Also, here Srivastava says "isomorphic" while elsewhere he says "Borel isomorphic."

One also wonders if there is a difference between a "Borel space" and a "standard Borel space."

There may also be something loaded into "topological space," since other definitions use the term "measurable space." We can chase down the definition of "topological space" to read about open sets and so on and so on.

  • 1
    $\begingroup$ A topology is completly metrizable if there is a complete metric on that space inducing the topology. $\endgroup$
    – dialegou
    May 15 at 16:05
  • 1
    $\begingroup$ Thanks, changed it. $\endgroup$
    – dialegou
    May 15 at 16:09
  • $\begingroup$ In particular, this doesn't mean that any metric inducing the topology is complete. $\endgroup$
    – dialegou
    May 15 at 16:10
  • $\begingroup$ "(Borel-)isomorphic to a separable complete metric space with the Borel σ-algebra" is equivalent to the definition in OP and I think the most common one. For example the Borel algebra on the rationals is Borel isomorphic to the Borel space of the discrete topology on a countable set and the latter is completely metrizable. $\endgroup$
    – dialegou
    May 15 at 16:44

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