Null Sets Closed under Countable Union/Signed Measure

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.

Let $\sequence {N_i}_{i \mathop \in \N}$ be a sequence of $\mu$-null sets.

Then:


 * $\ds N = \bigcup_{i \mathop = 1}^\infty N_i$

is a $\mu$-null set.

Proof
From Countable Union of Measurable Sets as Disjoint Union of Measurable Sets, there exists a sequence of pairwise disjoint $\Sigma$-measurable sets $\sequence {A_i}_{i \in \N}$ such that:


 * $\ds N = \bigcup_{i \mathop = 1}^\infty A_i$

We now show that if $E \in \Sigma$ has $E \subseteq N$, then $\map \mu E = 0$.

Write:

Since $\mu$ is countably additive, we have:


 * $\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i}$

We have, from Intersection is Subset:


 * $E \cap A_i \subseteq A_i \subseteq N_i$

Since $N_i$ is a $\mu$-null set, we have that:


 * $\map \mu {E \cap A_i} = 0$ for each $i$.

So:


 * $\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu {E \cap A_i} = 0$

So:


 * for each $E \in \Sigma$ with $E \subseteq N$, we have $\map \mu E = 0$.

So $N$ is a $\mu$-null set.