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When we are building some theory involving measures, there are cases where we should ignore null sets (sets of measure zero), and in some of those cases, we treat different sub sigma algebras only up to null sets. In this post, I list different characterizations of two sub sigma algebras being equal up to null sets, and related gotchas. This concerns the stage where you work with sub-sigma-algebras of a probability space before you move on to work with measure algebras (directly or indirectly).
Some quick definition before we start: two measurable sets and are equal mod 0 or equivalent mod 0 if their symmetric difference is a measure zero set.
1. closure of a sub-sigma-algebra
Given a sub sigma algebra (on a not necessarily complete probability space ), consider the following sets:
(1) where is the set of measure zero elements in .
(2) the set of elements in mod zero equivalent to an element in
(3) topological closure of w.r.t. the pseudo metric on defined by
(10) The sigma algebra associated with the completion of , i.e., the sigma algebra generated by and subsets of measure zero elements in .
- Prove that (1), (2), (3) are the same. I will call this set the closure of .1
- Under the assumption that the probability space is complete (i.e., on is a complete measure), prove that (10), which I will call the completion of , is a sub sigma algebra of (1).
- A gotcha. Under the same assumption, give an example of (10) being a proper subset of (1).2
Hints or answers are provided by footnotes at the bottom.
2. closed sub sigma algebra
Given a sub sigma algebra (on a not necessarily complete probability space ), consider following conditions:
(1′) it contains
(2′) it is closed under replacing with a mod zero equivalent, i.e., if and is mod zero equal to , then
(3′) Its closure is itself
(4′) it is closed under moving to an element of arbitrary close (in the sense of the pseudo metric ) to elements in , i.e., any element of arbitrarily close to elements in is in .
(10′) is a complete measure w.r.t. it.
- Prove that (1′), (2′), (3′), (4′) are equivalent conditions.
- A gotcha. Assuming complete probability space, prove that (1′) implies (10′) and that (1′) is actually strictly stronger than (10′).3
3. mod zero equivalence of sub sigma algebras
Given two sub sigma algebras , (on a not necessarily complete probability space ), consider the following conditions:
(1) they have the same closure
(2) given any set , is mod zero equal to an element in if and only if it is mod zero equal to an element in
(3) each element in one has a mod zero equivalent counterpart in the other
(10) they have the same completion
- Prove that (1), (2), (3) are equivalent conditions. When any of these three conditions hold, we say that the two sub sigma algebras are equivalent mod 0.
- Assuming complete probability space, prove that (10) implies (1).
- Assuming the same, give an example to show that (10) is strictly stronger than (1).4
Nevertheless, the sub sigma algebra and its closure and its completion are equivalent mod 0, so they are interchangeable in some sense, so don’t worry too much about the gotchas.
To prove that (3) is a subset of (2), recall a proof of Borel-Cantelli lemma
Take the probability space of Lebesgue measure on the unit square. The sub sigma algebra to take is the vertical sigma algebra, i.e., the sigma algebra generated by the first-coordinate canonical projection map. The diagonal in the unit square is in (1) but not in (10)
strictly stronger because of the aformentioned vertical sigma algebra.
Let be where give measure 1/2 to 1 and 2, but measure zero to 3. This is a standard probability space even. Let be the sub sigma algebra generated by and by . They are equivalent mod zero. Each is its own completion.