σ-algebra and measure of the union of 2 sets. Based on measure theory.

Question

A measure space $(\mathbb S,\mathcal S,μ )$ is not complete. The system of all its null sets is $\mathcal O$.

Let $$\mathcal S' = \{A ∪ O : A ∈ \mathcal S, O ∈ \mathcal O\}.$$ The formula of the function $μ' : \mathcal S' → [0,∞]$ is $$μ'(A ∪ O) = μ(A).$$

Questions:

(a) Show that $\mathcal S' = σ(\mathcal S ∪ \mathcal O).$

(b) Show: $μ'(A1 ∪ O1) = μ'(A2 ∪ O2)$, if $A1 ∪ O1 = A2 ∪ O2.$

My solution:

(a) I started from showing that the right side is σ-algebra:

1) $\mathbb S \in \mathcal S∪ \mathcal O?$

$\mathbb S \in \mathcal S.$ Then $\mathbb S ∪ \mathcal O = \mathbb S \in \mathcal S'.$

Answer 1

First show $\mathcal{S'}$ is a $\sigma$-algebra.

Note that $\mathcal{S} \subseteq \mathcal{S'}$ trivially (we can choose $O=\emptyset$) and $\mathcal{O} \subseteq \mathcal{S'}$ in the same way. Hence $\mathcal{S} \cup \mathcal{O} \subseteq \mathcal{S'}$.

Then $\sigma(\mathcal{S} \cup \mathcal{O}) \subseteq \mathcal{S'}$ follows by minimality.

The other inclusion is clear, because if $S \in \mathcal{S}$ and $O \in \mathcal{O}$, $S \cup O$ must lie in any ($\sigma$-)algebra that contains both $\mathcal{S}$ and $\mathcal{O}$, so in particular in $\sigma(\mathcal{S} \cup \mathcal{O})$.

To see that $\mathcal{S'}$ indeed is a $\sigma$-algebra:

• $\mathbb{S} = \mathbb{S} \cup \emptyset \in \mathcal{S'}$, as $\mathbb{S} \in \mathcal{S}$ and $\emptyset \in \mathcal{O}$.

• If $A \cup O \in \mathcal{S'}$ find $O' \in \mathcal{S}$ with $O \subseteq O'$ and $\mu(O') =0$, and then $(A \cup O)^\complement = A^\complement \cup O^\complement$ and

$$A^\complement \cap O^\complement = (A^\complement \cap O'^\complement) \cup (A^\complement \cap (O' \setminus O))$$

shows that the complement is in $\mathcal{S'}$, as the first part of the union is in $\mathcal{S}$ and the second part is a subset of the null set $O'$ so in $\mathcal{O}$.

• If $A_n \cup O_n$ is a sequence of sets in $\mathcal{S'}$, then its union equals $(\bigcup_n A_n) \cup (\bigcup_n O_n)$ which is in $\mathcal{S'}$ as the left part of the union is in $\mathcal{S}$ and a countable union of null sets is a null set.

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To show well-definedness of the extension $\mu'$ of $\mu$ you need to show that $A= A_1 \cup O_1 = A_2 \cup O_2$ in two ways, with $A_1,A_2 \in \mathcal{S}$ and $O_1,O_2 \in \mathcal{O}$, then $\mu(A_1)=\mu(A_2)$. Then we can indeed define $\mu'(A \cup O) = \mu(A)$ and the final fact follows.