lacunary - Mathnotes

Measurable Spaces and Functions

Theorem: nonexistence of extension of length to all subsets of R \@{nonexistence-of-extension-of-length-to-all-subsets-of-reals}

There does not exist a function μ with all of the following properties:

  1. (a) μ is a function from the set of subsets of R to [0,∞].
  2. (b) μ(I)=ℓ(I) for every @open interval I of R.
  3. (c) μ(⋃k=1∞Ak)=∑k=1∞μ(Ak) for every @disjoint sequence A1,A2,… of subsets of R.
  4. (d) μ(t+A)=μ(A) for every A⊂R and every t∈R.
Proof \@{proof-of-nonexistence-of-extension-of-length-to-all-subsets-of-reals}

Outline: assume such a μ exists. Then we can show that it has all of the properties of outer measure that were used to prove Nonadditivity of outer measure, but this means μ doesn't satisfy (c), a contradiction. TODO: full proof from paper notes.

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σ-Algebras

Example \@{example-4}
  • Suppose X is a set. Then {∅,X} is a σ-algebra on X.
  • Suppose X is a set. Then the @power set of X is a σ-algebra on X.
  • Suppose X is a set. Then the set of all subsets E of X such that E is countable or X∖E is countable is a σ-algebra on X.
Theorem: sigma algebras are closed under countable intersection \@{sigma-algebras-are-closed-under-countable-intersection}

Suppose S is a σ-algebra on a set X. Then

  1. (a) X∈S
  2. (b) if D,E∈S, then D∪E∈S and D∩E∈S and D∖E∈S
  3. (c) if E1,E2,… is a sequence of elements of S, then ⋂k=1∞Ek∈S.
Proof \@{proof-of-sigma-algebras-are-closed-under-countable-intersection}

TODO: full proof from paper notes.

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Example \@{example-8}

If X=R and S is the set of all subsets of R that are countable or have a countable complement, then Q is S-measurable but the set of positive real numbers is not S-measurable.

Theorem: smallest sigma algebra containing a collection of subsets. \@{smallest-sigma-algebra-containing-a-collection-of-subsets}

Suppose X is a set and A is a set of subsets of X. Then, the intersection of all σ-algebras on X that contain A is a σ-algebra on X.

Proof \@{proof-of-smallest-sigma-algebra-containing-a-collection-of-subsets}

We know there is at least one σ-algebra on X that contains A because the @power set of X is a @sigma-algebra and contains A.

Now, let S be the intersection of all σ-algebras on X that contain A. Then ∅∈S because by definition of σ-algebra, every σ-algebra contains the empty set. Suppose E∈S. Then E is in every σ-algebra on X that contains A, and by definition, each σ-algebra also contains X∖E, so X∖E∈S. Suppose E1,E2,… is a sequence of elements of S. Then each Ek is in every σ-algebra on X that contains A, and so is the union of these Ek, and so it is in their intersection S as well.

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Note \@{smallest-sigma-algebra-containing-a-collection-of-subsets-note}

Smallest makes sense for terminology here because this intersection is a subset of every σ-algebra that contains A.