Measurable Spaces – Problem (41/365)

Let \mathcal{D} = \{ D_1, D_2, \dots \} be a countable decomposition of \mathcal{\Omega} and \mathcal{A} = \sigma(\mathcal{D}) be the \sigma-algebra generated by \mathcal{D}. Are there only countably many sets in \mathcal{A}?

The answer is no. We can show this by showing that the natural numbers is a strict subset of \mathcal{A}. Every natural number n can be written as \sum_{i=0}^\infty a_i 2^i where a_i \in \{0,1\} and only a finite number of a_i = 1 (because if an infinite number of a_i = 1 the sum is \infty).

Note that since \mathcal{D} is a decomposition we can write every set in \mathcal{A} as a countable (because this is a \sigma-algebra) union of a subset of \mathcal{D}. This means we can encode every set in \mathcal{A} as \cup_{i=1}^\infty (A_i \cap D_i) where A_i \in \{ \emptyset, \Omega \}. First, since \mathcal{D} is countable there is a bijection between the natural numbers and \mathcal{D}, however, \mathcal{A} is not countable since we can have a countable number of A_i = \Omega.

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