Measure-Theoretic Lingo

Let \Omega be a set. A collection \mathcal F \subseteq \mathcal P(\Omega) is a \sigma-algebra on \Omega if it satisfies the following properties:

  • \emptyset, \Omega \in \mathcal F,
  • For any M \in \mathcal F, \Omega \backslash M \in \mathcal F.
  • For any M_1,M_2,\dots, \in \mathcal F, \displaystyle \bigcup_{i=1}^\infty X_i \in \mathcal F.

If such a \sigma-algebra exists, we call (\Omega,\mathcal F) a measurable space. Elements of \mathcal F are called measurable sets.

Example 1. For any set \Omega, (\Omega, \{\emptyset, \Omega\}) is a \sigma-algebra known as the trivial \sigma-algebra) and that (\Omega, \mathcal P(\Omega)) is a \sigma-algebra.

Problem 1. Let \Omega be a set and fix \mathcal K \subseteq \mathcal P(\Omega). Find a \sigma-algebra that contains \mathcal K, then define

\displaystyle \mathcal K' := \{\mathcal G \subseteq \mathcal P(\Omega) : \mathcal K \subseteq \mathcal G,\ \mathcal G\ \text{is a}\ \sigma\text{-algebra}\},\quad \sigma(\mathcal K) := \bigcap_{\mathcal G \in \mathcal K'}\mathcal G.

Prove that \sigma(\mathcal K) is the smallest \sigma-algebra that contains \mathcal K, called the \sigma-algebra generated by \mathcal K.

(Click for Solution)

Solution. Obviously, \mathcal K \subseteq \mathcal P(\Omega). For the first property, since \emptyset,\Omega \in \mathcal G for any \mathcal G \supseteq \mathcal K, \emptyset, \Omega \in \sigma(\mathcal K).

For the second property, fix M \in \sigma(\mathcal K). Then M \in \mathcal G for any \mathcal K \subseteq \mathcal G. Since each \mathcal G is a \sigma-algebra, \Omega \backslash M \in \mathcal G. Hence, \Omega \backslash M \in \sigma(\mathcal K).

For the third property, fix M_1,M_2,\dots \in \sigma(\mathcal K). Then for each i, M_i \in \mathcal G for any \mathcal K \subseteq \mathcal G. Since each \mathcal G is a \sigma-algebra, \bigcup_{i=1}^\infty M_i \in \mathcal G. Hence, \bigcup_{i=1}^\infty M_i \in \sigma(\mathcal K).

Problem 2. Let (\Omega, \mathcal F) be a measurable space. For any K \subseteq \Omega, define the collection \mathcal F_K \subseteq \mathcal P(K) of subsets by

\mathcal F_K := \{K \cap M : M \in \mathcal F\}.

Prove that (K, \mathcal F_K) is a measurable space.

(Click for Solution)

Solution. The first property is obvious. The second property follows from

K \backslash N = K \backslash (K \cap M) = K \backslash M = K \cap (\Omega \backslash M) \in \mathcal F_K.

The third property follows from

\displaystyle \bigcup_{i=1}^\infty (K \cap M_i) = K \cap \bigcup_{i=1}^\infty M_i \in \mathcal F_K.

—Joel Kindiak, 7 Feb 25, 2331H

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