Null Sets Closed under Countable Union

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Theorem

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

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


Then $N := \ds \bigcup_{n \mathop \in \N} N_n$ is also a $\mu$-null set.


Signed Measure

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.


Resolution of the Identity

Let $X$ be a topological space.

Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.

Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.

Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.

Let $\set {A_j : j \in \N} \subseteq \map \BB X$ such that:

$\map \EE {A_j} = 0$ for each $j \in \N$.

Let:

$\ds A = \bigcup_{j \mathop = 1}^\infty A_j$


Then $\map \EE A = 0$.


Proof

As $\mu$ is a measure:

$\map \mu N \ge 0$

Also:

\(\ds \map \mu N\) \(\le\) \(\ds \sum_{n \mathop \in \N} \map \mu {N_n}\) Measure is Countably Subadditive
\(\ds \) \(=\) \(\ds \sum_{n \mathop \in \N} 0\) as the $N_n$ are $\mu$-null sets
\(\ds \) \(=\) \(\ds 0\)

Hence necessarily:

$\map \mu N = 0$

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

$\blacksquare$


Sources

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