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Is it possible to place non-trivial probability measures on sets of cardinality strictly greater than the continuum -- in particular, on sets of cardinality 2^c? (Any references would be appreciated.)

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Yes, of course. You can always give $2^X$ the product measure. A decent reference is volume 2 ("Broad foundations") of Fremlin's treatise on Measure theory. Look at Section 254 (Infinite products). (The book may be available online from Fremlin's page.) – Andrés E. Caicedo Feb 28 '11 at 21:31
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There's even more. If your set can be considered as a product of probability spaces, its measure will have some good properties. This is a theorem due to Alexandra Ionescu Tulcea: given an infinite (with arbitrary cardinality) family of probability spaces $$\left\{(\Omega_n, \mathscr{B}_n, \mu_n)\right\}_{n \in I}$$ if we call $$\pi_m : \prod_{n \in I} \Omega_n \rightarrow \Omega_m$$ the canonical projection, then it's possible to give $\Omega \mathrel{\mathop{:}=} \prod_{n \in I} \Omega_n$ the structure of a probability space with measurable sets $$\mathscr{B} = \bigotimes_{n \in I} \mathscr{B}_n$$ and probability measure $\mu : \mathscr{B} \rightarrow \mathbb{R}$ such that $$\forall \{n_1, \ldots, n_t\} \subset I \ \ \forall (A_{n_1}, \ldots, A_{n_t}) \in \prod_{i = 1}^t \mathscr{B}_{n_i}$$ $$\mu\left(\pi_{n_1}^{-1}(A_{n_1}) \cap \ldots \cap \pi_{n_t}^{-1}(A_{n_t})\right) = \mu_{i_1}(A_{i_1}) \ldots \mu_{i_t}(A_{i_t})$$

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Isn't this just a special case of Kolmogorov's Consistency Theorem? – Noah Stein Feb 28 '11 at 22:46

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