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Taken from Condition Expectation in Quantum Probabilty by Denes Petz.

In quantum probability there are a number of fundamental questions that ask how faithfully can one quantise classical probability. Suppose that (\Omega,\mathcal{S},P) is a (classical) probability space and \mathcal{G}\subset\mathcal{A} a sub-\sigma-algebra. The conditional expectation of some integrable function f (with respect to some L-space) relative to \mathcal{G} is the orthogonal projection onto the closed subspace L(\mathcal{G}):

\mathbb{E}^{\mathcal{G}}:L(\mathcal{A})\rightarrow L(\mathcal{G})f\mapsto \mathbb{E}(f|\mathcal{G}).

Suppose now that (A,\rho) is a quantum probability space and that B is some C*-subalgebra of A. Can we always define a conditional expectation with respect to B? The answer turns out to be not always, although this paper gives sufficient conditions for the existence of such a projection. Briefly, things work the other way around. Distinguished states give rise to quantum conditional expectations — and these conditional expectations define a subalgebra. We can’t necessarily start with a subalgebra and find the state which gives rise to it — Theorem 2 gives necessary conditions in which this approach does work.

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Taken from Real Analysis and Probability by R.M. Dudley.

For a sequence of n repeated, independent trials of an experiment, some probability distributions and variables converge as n tends to infinity. In proving such limit theorems, it is useful to be able to construct a probability space on which a sequence of independent random variables is defined in a natural way; specifically, as coordinates for a countable Cartesian product.

The Cartesian product of finitely many \sigma-finite measure spaces gives a \sigma-finite measure space. For example, Cartesian products of Lesbesgue measure on the line give Lesbesgue measure on finite-dimensional Euclidean spaces. But suppose we take a measure space \{0,1\} with two points each having measure 1\mu(\{0\})=1=\mu(\{1\}), and form a countable Cartesian product of copies of this space, so that the measure of any countable product of sets equals the product of their measures. Then we would get an uncountable space in which all singletons have measure 1, giving the measure usually called counting measure. An uncountable set with counting measure is not a \sigma-finite space, although in this example it was a countable product of finite measure spaces. By contrast, the the countable product of probability measures will again be a probability space. Here are some definitions.

For each n=1,2,\dots let (\Omega_n,S_n,P_n) be a probability space. Let \Omega be the Cartesian product \displaystyle \prod_{n\geq 1}\Omega_n, that is, the set of all sequences \{\omega_n\}_{n\geq 1} with \omega_n\in\Omega_n for all n. Let \pi_n be the natural projection of \Omega onto \Omega_n for each n\pi_n\left(\{\omega_m\}_{m\geq 1}\right)=\omega_n for all n. Let S be the smallest \sigma-algebra of subsets of \Omega such that for all m\pi_m is measurable from (\Omega,S) to (\Omega_m,S_m). In other words, S is the smallest \sigma-algebra containing all sets \pi_n^{-1}(A) for all n and all A\in S_n.

Let \mathcal{R} be the collection of all sets \displaystyle \prod_{n\geq 1}A_n\subset\Omega where A_n\in \mathcal{S}_n for all n and A_m=\Omega_m except for at most finitely many values of n. Elements of \mathcal{R} will be called rectangles. Now recall the notion of semiring. \mathcal{R} has this property.

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