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As defined here http://modular.math.washington.edu/129/ant/html/node82.html

Using the notation in the link, one takes sets of the form $\prod\limits_{\lambda} M_{\lambda}$, where each $M_{\lambda}$ has finite measure and almost all of them have measure $1$. Then $\mu \prod\limits_{\lambda} M_{\lambda} = \prod\limits_{\lambda} \mu_{\lambda} M_{\lambda}$ is a finite product. What then, do we take the $\sigma$-algebra generated by sets of the above form and extend $\mu$ to a measure on them? It's not at all obvious how this can be done. Will we have to consider infinite convergent products?

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There is some serious measure theory going on in the background here. Firstly Dynkin's $\pi$-$\lambda$ theorem implies that if two finite measures agree on a $\pi$ system (set which is stable under finite intersections), then they must agree on the $\sigma$-algebra generated by that $\pi$ system. Sets of the form $\prod_\lambda M_\lambda$ are called measurable rectangles and they form a $\pi$ system. Thus, if $\Sigma$ is the $\sigma$-algebra generated by all such measurable rectangles, then there is at most one measure $\mu$ such that $\mu(\prod_\lambda M_\lambda) = \prod_\lambda \mu_\lambda(M_\lambda)$ for all measurable rectangles.

The next bit of serious measure theory is why should such a $\mu$ exist? The existence of such a measure is given by Kolmogorov's extension theorem. The theorem basically says that if you have something that looks like finite dimensional distributions of a stochastic process, then there actually is measure space whose coordinate process has the given finite dimensional distributions. In this case, the given products "look like finite dimensional distributions" because for finite products you can simply construct a product measure by hand.

I admit this is very wishy washy sounding, and that is because I am trying to explain in only a few words some very deep theorems that require a lot of background to prove. Any graduate course on measure theory or probability/stochastic processes will cover Dynkin's theorem, and maybe a weaker version of Kolmogorov's theorem for countable products called Ionescu-Tulcea, but few will actually prove Kolmogorov's extension theorem.

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  • $\begingroup$ Thanks for your response, I think I understand most of what you're saying. Would things be easier if the product was countable? (so I could use this Ionescu-Tulcea) Also, the only case I'm interested in is where the $X_{\lambda}$ are locally compact topological groups, and $Y_{\lambda}$ is a compact open subset of $X_{\lambda}$ (if you've seen ideles or adeles). In that case, one can argue that $\Omega$ is locally compact, so it must have a Haar measure. Would it help in any way to argue that the $\mu$ we're looking for is exactly that Haar measure? $\endgroup$
    – D_S
    Commented Mar 1, 2015 at 0:06
  • $\begingroup$ I'm not sure I understand what you are asking. In the case you specify, the content of the Ionescu-Tulcea theorem is exactly the existence of the measure $\mu$. I don't see any reason why this measure should be related to the Haar measure on $\Omega$, but I don't know much about Haar measures. $\endgroup$
    – nullUser
    Commented Mar 1, 2015 at 0:41
  • $\begingroup$ I think some confusion might be coming from the terminology: "basis" of measurable sets. The term basis here does not mean topological basis, it just means that the measurable rectangles generate the $\sigma$-algebra in question. $\endgroup$
    – nullUser
    Commented Mar 1, 2015 at 0:44

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