Does the infinite product of probability spaces always exist (using the sigma algebra that makes all projections measurable and providing a probability measure on this sigma algebra)?

I always assumed the answer was yes, but today I read in some lecture notes:

"...For that we will need to be able to construct infinite product measure spaces and extend natural measures to them. Interstingly the construction requires that the measure structure be tied to the natural topology on this space. A class of topologies that behave naturally under such construction goes under the name Polish; the measure spaces arising thereby are called standard Borel."

Can someone explain this paragraph to me? Was I mistaken about the general existence? What does topology have to do with it? Even if I was right, what exactly is special about standard Borel spaces when one is interested in questions that involve topology?


It seems that some of the confusion lives on in the answers and comments of others. After consulting the literature (I am not an expert in this area), I think I know what the confusion is about.

There are two common situations that arise when considering infinite families of probability measures. The first one concerns arbitrary products of probability spaces:

Situation 1. Let $I$ be a non-empty set, let $\{\Omega_i,\mathscr A_i,\mu_i\}_{i\in I}$ be an indexed family of probability spaces and let $(\Omega,\mathscr A)$ denote the product of their underlying measurable spaces. We want to find a probability measure $\mu$ on $(\Omega,\mathscr A)$ such that:

  • the coordinate projections $\pi_i : \Omega \to \Omega_i$ are independent random variables (with respect to $\mu$);
  • for each $i\in I$ the pushforward measure of $\mu$ along the coordinate projection $\pi_i : \Omega \to \Omega_i$ (in other words, the distribution of the random variable $\pi_i$) coincides with the original measure $\mu_i$.

Does such a probability measure exist? Is it unique?

A proposition in measure theoretic probability says that the answer to both questions is yes in all cases. Proofs of this theorem can be found in many of the more advanced textbooks on measure theory or probability. (See references below.) Note: though the proof uses only basic measure theory, it only works if the factors in the product are probability spaces. It does not extend to arbitrary measure spaces!

The second situation is adressed by Kolmogorov's extension theorem. Here the independence requirement is replaced by a much weaker consistency criterion.

Situation 2. Let $I$ be a non-empty set and let $\{\Omega_i,\mathscr A_i\}_{i\in I}$ be an indexed family of measurable spaces. For every non-empty $J \subseteq I$, let $(\Omega_J,\mathscr A_J)$ denote the partial product $\prod_{j\in J} (\Omega_j,\mathscr A_i)$ of measurable spaces. Furthermore, for $\varnothing \neq A \subseteq B \subseteq I$, we let $\pi^B_A : \Omega_B \to \Omega_A$ denote the coordinate projection (clearly $\pi^B_A$ is measurable). Finally, let $\mathscr F$ denote the collection of all finite, non-empty subsets of $I$.

Suppose that we have an indexed family of probability measures $\{\mu_F\}_{F\in \mathscr F}$ on the spaces $\{(\Omega_F,\mathscr A_F)\}_{F\in\mathscr F}$ with the following consistency property:

  • for all $A,B\in\mathscr F$ with $A \subseteq B$, the measure $\mu_A$ coincides with the pushforward measure of $\mu_B$ along $\pi^B_A$.

Does there exist a probability measure $\mu$ on $(\Omega,\mathscr A) := (\Omega_I,\mathscr A_I)$ such that for each $F\in\mathscr F$ the pushforward measure of $\mu$ along $\pi^I_F$ coincides with $\mu_F$? Is such a $\mu$ unique?

It turns out that the answer to this question is no in general (see Cohn, exercise 10.6.5). However, if the spaces $(\Omega_i,\mathscr A_i)$ are sufficiently nice, the answer is yes, and luckily this still covers many practical uses.

So what exactly is the difference between the two situations? To answer this question, you must convince yourself that the consistency requirement is much weaker than independence. In fact, situation 2 can be used to find a joint distribution for an infinite family of dependent random variables! This plays an important role in the theory of stochastic processes. (Indeed, Bauer proves the result in the first section of his chapter on stochastic processes; see references below.)

It is of course possible to use Kolmogorov's extension theorem in order to form infinite products of independent random variables: we simply specify the probability measure on every finite product $(\Omega_F,\mathscr A_F)$ to be the one that makes the random variables $\pi_i$ independent. It seems that the author of OP's lecture notes mistakenly thought that this is the only way to construct infinite sequences of independent random variables, and therefore incorrectly concluded that such sequences only exist in certain (topological) cases. An understandable mistake, but confusing nonetheless.

As a final word, those who have seen such constructions in other areas of mathematics may recognise situation 1 as a product of probability spaces and situation 2 as a projective limit of probability spaces. This nomenclature is also used by Bauer; see references below.


  • Donald L. Cohn, Measure Theory (Second Edition), Birkhäuser Advanced Texts: Basler Lehrbücher, 2013. Both theorems are treated in section 10.6, and it was this exposition that led me to understand the difference between the two results. The result from situation 1 is proved only for countable products in the main text; the general case is deferred to the exercises. Furthermore, an outline of a counterexample for the general case of situation 2 is given in the exercises.

  • Heinz Bauer, Robert R. Burckel (translator), Probability Theory, de Gruyter Studies in Mathematics 23, Walter de Gruyter, 1996. This is a very technical (but nevertheless great) introduction to probability theory from a measure theoretic point of view, which assumes knowledge of measure theory as a prerequisite. Situation 1 is covered in §9 (infinite products of probability spaces), and situation 2 is covered in §35 (projective limits of probability measures). The independence requirement in situation 1 is somewhat hidden in the wording of theorem 9.2, but it follows from the remarks preceding the theorem. An outline of a counterexample for the general case of situation 2 is given at the end of §35.

  • Paul R. Halmos, Measure Theory, Graduate Texts in Mathematics 18, Springer, 1974 (reprint of the 1950 edition by Van Nostrand). This book includes a clear proof and helpful remarks for situation 1 in §38 (infinite dimensional product spaces). Again the result is only proven for countable products, and the general case is deferred to the exercises. Like Bauer, he formulates the theorem without using the word independence. The book does not seem to treat situation 2.

  • Jacques Neveu supposedly also addresses situation 1 in his book Mathematical Foundations of the Calculus of Probability (translated from French), but I don't seem to have access to this book.

  • $\begingroup$ The perfect answer! Thank you so much for the clarification $\endgroup$ – Bananach Jul 18 '17 at 9:48
  • $\begingroup$ Situation 2 is the projective limit of the spaces $\left( \prod _{i \in F} \Omega_i, \prod _{i \in F} \mathscr A_i \right)_{F \in \mathscr F}$. $\endgroup$ – Alex M. Jul 18 '17 at 14:23
  • 1
    $\begingroup$ To summarize: the product probability that pushes forward to the given probabilities always exists; on the other hand, the projective limit probability of the projective system $\left( \prod _{i \in F} \mu_i \right)_{F \in \mathscr F}$ exists only under some topological assumptions. $\endgroup$ – Alex M. Jul 18 '17 at 14:28
  • $\begingroup$ @AlexM. Oh, I see, you knew this already (as demonstrated by this question of yours). I mistook your comment to one of the other answers for confusion on your side. Sorry! $\endgroup$ – Josse van Dobben de Bruyn Jul 18 '17 at 20:28

If you wish to construct product measure with infinitely many probability spaces as factors, then no topological assumptions are needed. More precisely, if $\{(E_t,\mathcal E_t,\Bbb P_t): t\in T\}$ is a non-empty collection of probability spaces indexed by some set $T$, then there is a unique probability measure $\Bbb P$ on the product space $(\times_{t\in T}E_t,\otimes_{t\in T}\mathcal E_t)$ such that the coordinate maps $\pi_s:\times_{t\in T}E_t\ni x=(x_t)_{t\in T}\to x_s\in E_s$, $s\in T$, are independent, and $\Bbb P(\pi_s^{-1}(B_s))=\Bbb P_s(B_s)$ for each $s\in T$. This is a consequence of a more general theorem of C. Ionescu Tulcea, and is discussed quite nicely by J. Neveu in Chapter V of his book Mathematical Foundations of the Calculus of Probability.


The problem is not to construct an infinite product of sigma-algebras. This always exists and is defined as you have stated, as the smallest sigma-algebra that makes all the projections measurable.

The problem is to find measures on this space. Of course a measure on the infinite product always pushes forward to a measure on the finite products contained in this space (since the projections are now measurable).

The issue is whether you can go back. That is: given the finite dimensional measures can you find a measure on the infinite product that pushes forward to these measures?

Under "topological" assumptions, such as asking that the space is Polish, the answer is yes, and given by Kolmogorov´s extension theorem: https://en.wikipedia.org/wiki/Kolmogorov_extension_theorem.

  • $\begingroup$ What if I want to take the $\mathbb {N} $ fold product of a probability space with itself? Does that always exist(including a probability measure, not just the sigma algebra on the product) without the assumption of polishness? $\endgroup$ – Bananach May 26 '16 at 11:08
  • $\begingroup$ @Bananach as is pointed out in the other answer, it depends on the probability measure you want to extend. A counterexample exists also with just countable products. It's treated here: people.hss.caltech.edu/~kcb/Notes/Kolmogorov.pdf. Something you might know, which could explain where you need the extra structure, is that for Polish spaces you have existence of Regular Conditional Probabilities. This is used in the Proof of Ionescu Tulcea. $\endgroup$ – Kore-N May 26 '16 at 20:01
  • $\begingroup$ Do you know why the result mentioned by @JohnDawkins in his answer makes absolutely no use of topological conditions (or, more generally, inner regularity with respect to some compact class)? I am talking about Proposition V.1.2 on page 166 of "Mathematical Foundations of the Calculus of Probability" by Jacques Neveu. $\endgroup$ – Alex M. Sep 11 '16 at 14:04
  • $\begingroup$ @AlexM. your question seems to be in the same vein as (if not identical to) the original question, which led me to add an answer that hopefully clarifies it a bit more. This comment serves to notify you. :-) $\endgroup$ – Josse van Dobben de Bruyn Jul 18 '17 at 13:18

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