Can the set of singletons with positive measure in a probability space be uncountable?
Attempt: I think the answer is no. I'm guessing that if this were true it would contradict the fact that the probability of the whole space is 1. How can I prove this? The probability of the whole space is the expectation (i.e. the integral) of the indicator function of the whole space. So I think I need to derive that the integral of the probability space is greater than 1 (which would be a contradiction). At this point, I need to exploit the uncountability hypothesis, but I don't see how to do that. Presumably I would simple functions somehow.
Best Answer
This boils down to the fact that the sum of uncountably many positive reals is infinite. One can see that in the following way: let $A$ be the uncountable set of positive reals we are adding. Let $A_i = \{a\in A : a>2^{-i}\}$ for $i\in \mathbb{N}$. If any $A_i$ is infinite, then the desired sum is infinite. (As noted below, one has to worry about the measurability of $A_i$. This can be ensured by taking a countable subset of $A_i$ and using the countable additivity of the measure.) Thus $A$ is the union of countably many finite sets, and so is countable, contradicting the supposition.