TheoremBase

Null Sets and Properties Holding Almost Everywhere

A measurable set is null exactly when its measure is zero; subsets and finite unions of null sets are null, and so is a countable union of sets each contained in a member of a given sequence of measurable sets of measure zero. Properties holding almost everywhere are preserved under enlargement and under finite intersection.

Statement

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let (X,F,μ)(X,\mathcal{F},\mu) be a measure space, so that (X,F)(X,\mathcal{F}) is a measurable space and μ\mu is a measure on it, with values in the set [0,∞][0,\infty] of The Extended Real Line, Its Order, and [0,∞] (ii). Let P(X)\mathcal{P}(X) be the power set of XX, let unions, intersections and differences of classes be those of The Boolean Operations on Classes, Disjointness, and the Universal Class (i), let ⊆\subseteq be the subclass relation, and, for a sequence (Am)(A_{m}), let ⋃m∈NAm\bigcup_{m\in\mathbb{N}}A_{m} be the union of (Am)(A_{m}) as a family indexed by N\mathbb{N}. A subset of XX is μ\mu-null as in Null Set of a Measure (i), and a class holds μ\mu-almost everywhere (μ\mu-a.e.) as in A Property Holding Almost Everywhere (i).

Let B∈FB\in\mathcal{F}. Then BB is μ\mu-null if and only if μ(B)=0\mu(B)=0.

Let NN and MM be sets with N⊆M⊆XN\subseteq M\subseteq X. If MM is μ\mu-null, then NN is μ\mu-null.

If NN and MM are μ\mu-null subsets of XX, then N∪MN\cup M is μ\mu-null.

Let (Bm)(B_{m}) be a sequence in F\mathcal{F} with μ(Bm)=0\mu(B_{m})=0 for every m∈Nm\in\mathbb{N}. Then μ(⋃m∈NBm)=0\mu\bigl(\bigcup_{m\in\mathbb{N}}B_{m}\bigr)=0, where ⋃m∈NBm∈F\bigcup_{m\in\mathbb{N}}B_{m}\in\mathcal{F} by σ-Algebra and Measurable Space (i).

Let (Bm)(B_{m}) be a sequence in F\mathcal{F} with μ(Bm)=0\mu(B_{m})=0 for every m∈Nm\in\mathbb{N}, and let (Nm)(N_{m}) be a sequence in P(X)\mathcal{P}(X) with Nm⊆BmN_{m}\subseteq B_{m} for every m∈Nm\in\mathbb{N}. Then ⋃m∈NNm\bigcup_{m\in\mathbb{N}}N_{m} is μ\mu-null.

Let QQ and RR be classes.

If QQ holds μ\mu-a.e. and X∩Q⊆RX\cap Q\subseteq R, then RR holds μ\mu-a.e.

If QQ and RR hold μ\mu-a.e., then Q∩RQ\cap R holds μ\mu-a.e.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…