Consider a (nonnegative) measure space , where is bounded. Scaling , we may assume without loss of generality that is a probability space. If is a measurable function, then there’s a cute proof that is almost surely (a.s.) constant on the atoms of . Let be an atom, and define a constant . We will show that a.s. on . Define measurable sets and . If both have measure zero, then we’re done. Otherwise, it suffices by symmetry to consider the case . Then , so
which is a contradiction. By proving that the real and imaginary parts of a complex-valued measurable function are a.s. constant, we can extend the preceding result to the complex case. In fact, by proving that each of the coordinate projection mappings are constant, the proof method works for finite-dimensional complex normed spaces. I don’t know to whom this proof should be attributed, but I encountered the one-dimensional real case here on Math.StackExchange.
In probability theory, in particular stochastic processes, we often study measurable functions on probability space into a separable metric space equipped with the Borel -algebra. For example, Brownian motion is a measurable function between a probability space and the space of continuous functions on equipped with the -metric. Although said metric spaces may come with an additional structure, such as being a Banach space, in which case it makes sense to define integration, they need not. So the first proof, however cute it might be, does not generalize well.
To get some intuition for a new argument, suppose is a simple function, where are pairwise disjoint measurable subsets covering , taking values in the metric space . I realize that does not have a vector space structure and the notation doesn’t make mathematical sense, but it is convenient. Let be an atom. Since , we must have for some . Hence, a.s. on .
Now suppose that a measurable function is the a.s. pointwise limit of simple functions where form a measurable partition of . For , there is a null set such that for every . is a null set, and taking the union with another null set, if necessary, we may assume that surely on . Hence,
So if we can show that a measurable function is the a.s. pointwise limit of simple functions, we will be done.
Lemma 1 Suppose that is a finite measure space and is a separable metric space. If is -measurable, then is the a.s. pointwise limit of simple functions.
Proof: Let be a countable dense subset of . For , let denote the open ball of radius centered at , and let be the union over all . It is clear that and therefore . For , set and inductively define . Observe that is a partition of into measurable subsets. For each , let be an integer such that
so that .
Fix an element , and define a sequence of simple functions by
If , then there is some such that
For given, we can find such that . Then for each ,
for some . To complete the proof, we observe that and therefore a.s.
We summarize our results in the following proposition.
Proposition Let be a probability space, and let be a separable metric space. A measurable function is a.s. constant on any -atom of .
I have not been able to get very far on necessary conditions on for measurable to be a.s. constant on atoms. I would like to be able to drop the explicit topological conditions, and an instead consider a measure space . But so far I only know that a measurable function need not be a.s. constant on atoms if is not countably generated, even if still separates points of . Nate Elredge provided the following elementary example in the MathOverflow post thread “Are measurable functions almost surely constant on atoms”.
Let be an uncountable set, and let be the -algebra of countable and cocountable subsets of (see Lemma 3). Define a nonnegative set function by
(Note that for any binary function , is completely determined by whether is countable or cocountable, then , if is countable, since a measure, by definition, vanishes on the empty set.).
Let be a countable collection of pairwise disjoint sets in . We partition the indices by defining and to be the subsets of countable and cocountable indices, respectively. To show that is -additive and therefore a probability measure, we need to show that is cocountable for at most one index . Indeed,
Suppose is an index such that is cocountable. Since is uncountable and is necessarily countable, is necessarily uncountable. For any , , whch implies that is countable.
Every cocountable set is an atom. Indeed, let be a cocountable set. If is measurable with , then is cocountable, equivalently . Let be a bijective function. For , is (co-) countable if and only if is (co-) countable, so is measurable. Bijectivity means that is not constant, in particular on any cocountable set.
Proof: Observe that , where is countable.
Proof: It is evident that and is closed under relative complement. Let be a collection of pairwise disjoint sets in . We can split the indices into disjoint and , where is countable and is cocountable. is countable and is cocountable. From Lemma 2, it follows that is cocountable.