Why should a physicist care about measurability of random variables?

Question

I'm working with some physicists at the moment, and I made the remark that random variables are, technically, defined as measurable functions on some "background" probability space, hence requiring the choice of two sets and two sigma algebras. They immediately asked where the term "measurable" came from (I gave the brief history about measuring lengths and areas), and then asked if it had anything to do, in general, with whether or not something was "measurable in a laboratory" (I said I wasn't sure). They then proceeded to say that it was nonsensical to consider any "non-measurable" random variables because such things are, almost by definition, not "interesting" to the physicist.

So, how can I explain to my physicist colleagues that we actually do care about measurability? Is there an example of a physical quantity which would otherwise be interesting but which turns out to be non-measurable (in the mathematical sense), and which might justify my mathematical nitpickiness to them? I'm OK with examples from ordinary probability or stochastic process theory.

Answer 1

This is easily explained

Suppose there are three states of the world a, b c. And suppose we receive a signal $s$ from $\{a,b,c\}$ to $\mathbb{R}$ such that $s(a)= s(b)> s(c)$.

A random variable is measurable on the algebra generated by $s$ is precisely those that can't distinguish between states $A$ and $b$. So measurability is what you can measure in a lab if what you are measuring depends exclusively on the data given by the signal generating the $\sigma$-algebra.

When there is a finite number of states in a measure space $\Omega$ every algebra of events is defined by a signal $s\colon \Omega\to \mathbb{R}$. A a random variable $f\colon \Omega \to \mathbb{R}$ is measurable only if $f= g\circ s$ for some function $g\colon \mathbb{R} \to \mathbb{R}$. So measirability is precisely measurability based on available data.

With an infinite state space, say $[0,1]$ an event $E\subset [0,1]$ is non-measurable if it is inconceivable for you to describe how you would distinguish in any conceivable laboratory if a state lies in $E$ or its compliment when the signal you receive is $s(a) = a$ for all $a\in [0,1]$. You cannot conceivably imagine an experiment that can answer this question.

Of course, if the physicist wants to integrate a function that is not measurable; then there is no way of doing this for the reasons above.

Answer 2

I don't think can explain that they should care about measurability.

The reason is that the question about existence or non-existence of non-measurable sets is a deep theoretical issue. It depends on how you define the underlying rules of set theory. To be more precise: There exists models of set theory (according to the ZF axioms) where all sets are measurable. (In these models there are other pathologies...)

In other words: Whether there are non-measurable sets or not depends on the chosen foundations of set theory. As far as I know no physical experiment is know to decide which foundations are the right ones...

Answer 3

$X$ being $\mathcal{F}$-measurable basically means that if you know the answer to the question "did $A$ happen?" for all $A \in \mathcal{F}$, then you can deduce the value of $X$. This is precisely correct in probability spaces where all elements have positive probability. In a space like $[0,1]$ with the usual uniform distribution, this is not quite right (because in this space all random variables would be measurable with the definition above, since the $\sigma$-algebra includes singletons).

A better formulation is that you can infer the value of $X$ given the answers to the question "did $A$ happen?" for all $A \in \mathcal{F}$ with positive probability. When this distinction is relevant, it typically has some physical content, loosely speaking because only sets of positive probability are "big enough" that we can determine whether we are in them using only "finite information". In that regard measurability is similar to continuity. (Indeed, a subset of the power set of a finite set is a $\sigma$-algebra if and only if it is a topology, so on finite domains measurability and continuity are the same.)

Another view is perhaps a bit strange from the mathematician's point of view, which is that one $\sigma$-algebra is not so interesting but a whole bunch of them is interesting. A family of $\sigma$-algebras $\mathcal{F}_t$ is called a filtration if $\mathcal{F}_s \subset \mathcal{F}_t$ when $s \leq t$. There is a natural way of generating a filtration from a stochastic process. The elements of the filtration at a particular time $t$ are exactly the events $A$ with the property that you can know whether they happened by only observing the process up to time $t$. This concept of "growing information" is quite physically natural, and is closely related to conditional expectation. It's unfortunate that the formalism is a bit artificial.

What we should not expect physicists to care about is situations where a $\sigma$-algebra is "big" (e.g. the Lebesgue $\sigma$-algebra on $\mathbb{R}$) but nonmeasurable sets/functions exist anyway. Generically these matters are sensitive to your choice of foundations, which is unacceptable to a physicist. The one exception I can think of is that white noise cannot exist. Loosely speaking, this happens because although our $\sigma$-algebra for general functions is big, the space itself is so overwhelmingly huge that the $\sigma$-algebra is small by comparison.

Answer 4

I can’t explicitly give you an example of where non-measurability IS relevant physically, but I can suggest a reason why the idea would not be intuitive to a physicist: consider the Banach-Tarski paradox. It is styled as a paradox exactly because the results are not physical (i.e sensible to a physicist); the necessary partitioning operations cannot be performed on actual matter, only on non-measurable sets.