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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