## Measurable Functions that Are Almost Surely Constant on Atoms

Consider a (nonnegative) measure space ${(\Omega,\mathcal{A},\mu)}$, where ${\mu}$ is bounded. Scaling ${\mu}$, we may assume without loss of generality that ${(\Omega,\mathcal{A},\mu)}$ is a probability space. If ${f:\Omega\rightarrow\mathbb{R}}$ is a measurable function, then there’s a cute proof that ${f}$ is almost surely (a.s.) constant on the atoms of ${\Omega}$. Let ${A\in\mathcal{A}}$ be an atom, and define a constant ${k:=\mu(A)^{-1}\int_{A}fd\mu}$. We will show that ${f=k}$ a.s. on ${A}$. Define measurable sets ${B_{1}:=\left\{f and ${B_{2}:=\left\{f>k\right\}\cap A}$. If both ${B_{1},B_{2}}$ have measure zero, then we’re done. Otherwise, it suffices by symmetry to consider the case ${\mu(B_{1})>0}$. Then ${\mu(B_{1})=\mu(A)}$, so

$\displaystyle k=\dfrac{1}{\mu(A)}\int_{A}fd\mu=\dfrac{1}{\mu(A)}\int_{B_{1}}fd\mu

which is a contradiction. By proving that the real and imaginary parts of a complex-valued measurable function ${f:\Omega\rightarrow\mathbb{C}}$ 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 ${\sigma}$-algebra. For example, Brownian motion is a measurable function between a probability space and the space ${(C[0,1], \left\|\cdot\right\|_{\infty})}$ of continuous functions on ${[0,1]}$ equipped with the ${\infty}$-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 ${f=\sum_{i=1}^{n}y_{i}1_{E_{i}}}$ is a simple function, where ${E_{1},\ldots,E_{n}}$ are pairwise disjoint measurable subsets covering ${\Omega}$, taking values in the metric space ${(Y,d)}$. I realize that ${Y}$ does not have a vector space structure and the notation doesn’t make mathematical sense, but it is convenient. Let ${A\subset \Omega}$ be an atom. Since ${0<\mu(A)=\sum_{i=1}^{n}\mu(A\cap E_{i})}$, we must have ${\mu(A\cap E_{i})=\mu(A)}$ for some ${1\leq i\leq n}$. Hence, ${f=y_{i}}$ a.s. on ${A}$.

Now suppose that a measurable function ${f}$ is the a.s. pointwise limit of simple functions ${f_{n}=\sum_{i=1}^{m_{n}}y_{n,i}E_{n,i}}$ where ${E_{n,1},\ldots,E_{n,m_{n}}}$ form a measurable partition of ${\Omega}$. For ${n\geq 1}$, there is a null set ${N_{n}}$ such that ${f_{n}(\omega)=y_{n}\in Y}$ for every ${\omega\in N_{n}}$. ${N:=\bigcup_{n=1}^{\infty}N_{n}}$ is a null set, and taking the union with another null set, if necessary, we may assume that ${f_{n}\rightarrow f}$ surely on ${N}$. Hence,

$\displaystyle f(\omega)=\lim_{n\rightarrow\infty}f_{n}(\omega)=\lim_{n\rightarrow\infty}y_{n}\ \forall \omega \in N$

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 ${(\Omega,\mathcal{A},\mu)}$ is a finite measure space and ${(Y,d)}$ is a separable metric space. If ${f:\Omega\rightarrow Y}$ is ${(\mathcal{A},\mathcal{B}(Y))}$-measurable, then ${f}$ is the a.s. pointwise limit of simple functions.

Proof: Let ${\left\{y_{i}\right\}_{i=1}^{\infty}}$ be a countable dense subset of ${Y}$. For ${n\geq 1}$, let ${B_{n}(y_{i})}$ denote the open ball of radius ${1/n}$ centered at ${y_{i}}$, and let ${B_{n}}$ be the union over all ${i}$. It is clear that ${B_{n}=Y}$ and therefore ${f^{-1}(B_{n})=\Omega}$. For ${n\geq 1}$, set ${A_{n,1}:=f^{-1}(B_{n}(y_{1}))}$ and inductively define ${A_{n,k}:=f^{-1}(B_{n}(y_{k}))\setminus\bigcup_{i=1}^{k-1}A_{n,i}}$. Observe that ${\left\{A_{n,i}\right\}_{i=1}^{\infty}}$ is a partition of ${\Omega}$ into measurable subsets. For each ${n}$, let ${i_{n}}$ be an integer such that

$\displaystyle \mu\left(C_{n}:=\Omega\setminus\bigcup_{i=1}^{i_{n}}A_{n,i}\right)<\dfrac{1}{2^{n}}$

so that ${\mu(D_{n}:=\bigcup_{m\geq n}C_{m})<2^{-n+1}}$.

Fix an element ${y_{0}\in Y}$, and define a sequence of simple functions by

$\displaystyle f_{n}:=y_{0}1_{D_{n}}+\sum_{i=1}^{i_{n}}y_{i}1_{A_{n,i}},\ \forall n\geq 1$

If ${\omega\notin\bigcap_{n=1}^{\infty}D_{n}}$, then there is some ${n}$ such that

$\displaystyle \omega\notin D_{n}\Rightarrow\omega\notin C_{m}\Leftrightarrow\omega\in\bigcup_{i=1}^{i_{m}}A_{m,i}\ \forall m\geq n$

For ${\varepsilon>0}$ given, we can find ${m_{0}\geq n}$ such that ${m_{0}^{-1}<\varepsilon}$. Then for each ${m\geq m_{0}}$,

$\displaystyle \left|f_{m}(\omega)-f(\omega)\right|=\left|y_{i}(\omega)-f(\omega)\right|

for some ${1\leq i\leq i_{m}}$. To complete the proof, we observe that ${\mu(\bigcap_{n=1}^{\infty}D_{n})=0}$ and therefore ${f_{n}\rightarrow f}$ a.s. $\Box$

We summarize our results in the following proposition.

Proposition Let ${(\Omega,\mathcal{A},\mu)}$ be a probability space, and let ${(Y,d)}$ be a separable metric space. A measurable function ${f:\Omega\rightarrow Y}$ is a.s. constant on any ${\mu}$-atom of ${\mathcal{A}}$.

I have not been able to get very far on necessary conditions on ${Y}$ for measurable ${f:\Omega\rightarrow Y}$ 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 ${(Y,\mathcal{Y})}$. But so far I only know that a measurable function ${f:\Omega\rightarrow Y}$ need not be a.s. constant on atoms if ${\mathcal{Y}}$ is not countably generated, even if ${\mathcal{Y}}$ still separates points of ${Y}$. Nate Elredge provided the following elementary example in the MathOverflow post thread “Are measurable functions almost surely constant on atoms”.

Let ${\Omega}$ be an uncountable set, and let ${\mathcal{A}}$ be the ${\sigma}$-algebra of countable and cocountable subsets of ${\Omega}$ (see Lemma 3). Define a nonnegative set function ${\mu:\mathcal{A}\rightarrow\left\{0,1\right\}}$ by

$\displaystyle\mu(A):=\begin{cases} {1} & {A\text{ is cocountable}}\\ {0} & {A\text{ is countable}}\end{cases}$

(Note that for any binary function ${\mu}$, is completely determined by whether ${A}$ is countable or cocountable, then ${\mu(A)=0}$, if ${A}$ is countable, since a measure, by definition, vanishes on the empty set.).

Let ${\left\{A_{n}\right\}_{n=1}^{\infty}}$ be a countable collection of pairwise disjoint sets in ${\mathcal{A}}$. We partition the indices ${\mathbb{N}}$ by defining ${I_{1}}$ and ${I_{2}}$ to be the subsets of countable and cocountable indices, respectively. To show that ${\mu}$ is ${\sigma}$-additive and therefore a probability measure, we need to show that ${A_{n}}$ is cocountable for at most one index ${n}$. Indeed,

$\displaystyle \mu\left(\bigcup_{m=1}^{\infty}A_{m}\right)=1=\sum_{m=1}^{\infty}\mu(A_{m})\Leftrightarrow \sum_{m\in I_{2}}\mu(A_{m})=1$

Suppose ${n}$ is an index such that ${A_{n}}$ is cocountable. Since ${\Omega}$ is uncountable and ${A_{n}^{c}}$ is necessarily countable, ${A_{n}}$ is necessarily uncountable. For any ${m\neq n}$, ${A_{m}\subset A_{n}^{c}}$, whch implies that ${A_{m}}$ is countable.

Every cocountable set is an atom. Indeed, let ${A\in\mathcal{A}}$ be a cocountable set. If ${B\subset A}$ is measurable with ${\mu(B)>0}$, then ${B}$ is cocountable, equivalently ${\mu(B)=1}$. Let ${f:\Omega\rightarrow\Omega}$ be a bijective function. For ${A\in\mathcal{A}}$, ${f^{-1}(A)}$ is (co-) countable if and only if ${A}$ is (co-) countable, so ${f}$ is measurable. Bijectivity means that ${f}$ is not constant, in particular on any cocountable set.

Lemma 2 If ${A\subset\Omega}$ is countable and ${B\subset\Omega}$ is cocountable, then ${A\cup B}$ is cocountable.

Proof: Observe that ${\Omega\setminus(A\cup B)=A^{c}\cup B^{c}\subset B^{c}}$, where ${B^{c}}$ is countable. $\Box$

Lemma 3 Let ${\Omega}$ be an infinite set, and let ${\mathcal{A}}$ be the collection of all countable and cocountable subsets of ${\Omega}$. Then ${(\Omega,\mathcal{A})}$ is a measurable space.

Proof: It is evident that ${\emptyset,\Omega\in\mathcal{A}}$ and ${\mathcal{A}}$ is closed under relative complement. Let ${\left\{A_{n}\right\}_{n=1}^{\infty}}$ be a collection of pairwise disjoint sets in ${\mathcal{A}}$. We can split the indices into disjoint ${I_{1}}$ and ${I_{2}}$, where ${n\in I_{1}\Leftrightarrow A_{n}}$ is countable and ${n\in I_{2}\Leftrightarrow A_{n}}$ is cocountable. ${\bigcup_{n\in I_{1}}A_{n}}$ is countable and ${\Omega\setminus\bigcup_{n\in I_{2}}=\bigcap_{n\in I_{2}}A_{n}^{c}\Leftrightarrow \bigcup_{n\in I_{2}}A_{n}}$ is cocountable. From Lemma 2, it follows that ${\bigcup_{n}A_{n}}$ is cocountable. $\Box$