Null Sets Closed under Countable Union/Signed Measure
Jump to navigation
Jump to search
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:
\(\ds E\) | \(=\) | \(\ds E \cap N\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{i \mathop = 1}^\infty \paren {E \cap A_i}\) | Intersection Distributes over Union of Family |
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.
$\blacksquare$