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Probability: Theory and Examples 01

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Measure Theory

Probability Spaces

Here and throughout the book, terms being defined are set in boldface.

Here and in what follows, countable means finite or countably infinite.

\(\sigma\)-field(or \(\sigma\)-algebra) \(\mathcal{F}\) is a (nonempty) collection of subsets of \(\Omega\) that satisfy

(i) if \(A \in \mathcal{F}\) then \(A^c \in \mathcal{F}\), and

(ii) if \(A_i \in \mathcal{F}\) is a countable sequence of sets then \(\cup_i A_i \in \mathcal{F}\).

Since \(\cap_i A_i = (\cup_i A_i^c)^c\), it follows that a \(\sigma\)-field is closed under countable intersections.

\((\Omega, \mathcal{F})\) is called a measurable space, i.e., it is a space on which we can put a measure.

A measure is a nonnegative countably additive set function; that is, a function \(\mu: \mathcal{F} \to \mathbf{R}\) with

(i) \(\mu(A) \ge \mu(\emptyset)\) for all \(A \in \mathcal{F}\), and

(ii) if \(A_i \in \mathcal{F}\) is a countable sequence of disjoint sets, then

\[\mu(\cup_i A_i) = \sum_{i}\mu(A_i)\]

If \(\mu(\Omega) = 1\), we call \(\mu\) a probability measure. In this book, probability measures are usually denoted by \(P\).

Theorem 1.1.1. Let \(\mu\) be a measure on \((\Omega,\mathcal{F})\)

(i) monotonicity. If \(A \subset B\) then \(\mu(A) \le \mu(B)\).

(ii) subadditivity. If \(A \subset \cup_{m=1}^{\infty} A_m\) then \(\mu(A) \le \sum_{m=1}^{\infty} \mu(A_m)\).

(iii) continuity from below. If \(A_i \uparrow A\) (i.e., \(A_1 \subset A_2 \subset \dots\) and \(\cup_i A_i = A\)) then \(\mu(A_i) \uparrow \mu(A)\).

(iv) continuity from above. If \(A_i \downarrow A\) (i.e., \(A_1 \supset A_2 \supset \dots\) and \(\cap_i A_i = A\)), with \(\mu(A_1) < \infty\) then \(\mu(A_i) \downarrow \mu(A)\).

Proof. (i) Let \(B - A = B \cap A^c\) be the difference of the two sets. Using \(+\) to denote disjoint union, \(B = A + (B-A)\) so

\[\mu(B) = \mu(A) + \mu(B-A) \ge \mu(A)\]

Lemma. Let \(I \ne \emptyset\) is an arbitrary index set (i.e., possibly uncountable). If \(\mathcal{F}_i, i \in I\) are \(\sigma\)-fieldm then \(\cap_{i\in I} \mathcal{F}_i\) is \(\sigma\)-field.

Proof. (i) If \(A \in \cap_{i \in I} \mathcal{F}_i\), then \(A \in \mathcal{F}_i, \forall i\). It follows \(A^c \in \mathcal{F}_i, \forall i\), and \(A^c \in \cap_{i\in I} \mathcal{F}_i\).

(ii) Similarily if \(A_j \in \cap_{i\in I}\mathcal{F}\) is a countable sequence of sets, then \(\cup_{j} A_j \in \cap_{i \in I} \mathcal{F}\).

Frorm this it follows that if we are given a set \(\Omega\) and a collection \(\mathcal{A}\) of subsets of \(\Omega\), then there is a smallest \(\sigma\)-field containing \(\mathcal{A}\). We will call this the \(\sigma\)-field generated by \(\mathcal{A}\) and denote it by \(\sigma(A)\).

Let \(\mathcf{R}^d\) btthe set of vectors \((x_1,\dots,x_d)\) of real numbers

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