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I was reading an article on weak convergence which described the theory via exercises. The article can be found here. Check out exercise $2.1$. I came across the following exercise which I could not completely solve:

Let $(X,\mathcal{B}(X),\mathbb{P})$ be a probability space where $X$ is a topological space that is uncountable and $\mathcal{B}(X)$ is the Borel sigma algebra on $X$.

Prove that there exists a countable family of disjoint sets such that each of those sets has a strictly positive $\mathbb{P}$ measure.

What I tried:

Define $\mathcal{S} =$collection of all possible countable measurable partitions of X. Note that each element of $\mathcal{S}$ is a sequence of disjoint sets.

Now of all the sequences, if either of them have a countable subcollection of sets of strictly positive measure, I am done. So the bad case would be if in $\mathcal{S}$ "every countable partition has only finite number of sets with strictly positive measure". Unfortunately I could not get a proof from here. I feel that I would get a contradiction by arguing that if the bad case happened, then $X$ cannot be uncountable.

I would be grateful for any hints to achieve this. If this is not the way to proceed and there is another way, I'd like to hear about it. Feel free to request clarifications.

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up vote 2 down vote accepted

I don't think the claim is correct:

if we choose $X = [0,1]$ and define $$\displaystyle \mathbb{P}\left[(a,b)\right] = \mathbb{1}_{0.5 \in (a,b)}$$

Here $\mathbb{1}_{0.5 \in (a,b)}$ stands for the indicator function.

For this example, any partition of $X$ can have at most one set that contains $0.5$ and therefore at most one set of a partition can have positive measure!

I think we need more conditions on the probability measure $\mathbb{P}$

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This is indeed the case. Could you kindly read pg 1 of the article (if you haven't done so) and tell me if there is any error in what I asked and what they asked? – Gautam Shenoy Dec 14 '12 at 8:44

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