# Why do we want maps to be measurable (in countably-additive setting)

When I have to explain things that I am doing to people who did not do (or even did not learn) measure-theoretical probability, I think of getting a question in the title, and I am not sure I have arguments strong enough to convince that measurability is indeed the must.

Let me focus on a particular case of stochastic optimization in discrete time, in the setting of countably-additive probabilities. Being it formulated as a dynamic programming, or a gambling problem, one requires the decision to be Borel (universally, analytically) measurable with respect to the current state. Such requirement further leads to known issues as non-existence of measurable selectors. I do understand, that if the selection is not measurable, then it is not possible to define a probability measure over the space of state trajectories in a formal way. However I am no sure whether such explanation is convincing enough, since one can further argue that formal definition of probability requires measurability just in this particular framework, and perhaps the framework is something to be fixed.

It seems that Dubins and Savage used the finitely-additive framework at least for the reason to escape the measurability issues (as it written in Section 1.3 "Gambles"). So does it mean, that this question can be reduced to whether to deal with $\sigma$-additive probability or finitely-additive one?

Edited: in the edit I wanted to address the point raised by Yuri in his answer. I do know that measurability serves at least for the two goals: to define the regularity (nice sets, nice functions) and the information/dependence structure which follows from the following theorem:

If $X$ and $Y$ are measurable maps such that $Y$ is $X$-measurable, then there exists a measurable map $f$ such that $Y = f(X)$.

Note, however, that $X$-measurability of $Y$ guarantees the existence of a measurable map $f$. If we are just interested in a non-meaurable one, then a necessary and sufficient condition would be that level sets (pre-images) of $Y$ are saturated w.r.t. level sets of $X$. As far as I know, the dependence/information structures e.g. in game theory may be defined just based on such partitions, and serve equally well for that goal as much as $\sigma$-algebras. On the other hand, the use of the latter comes with an expense of dealing with measurable maps exclusively, which appears to be unnecessary at least for the particular of defining information structures.

The regularity task seems to be more in the need of measurability, but the only reason I could see here is the ability to measure nice sets or, more generally, to integrate nice functions. Indeed, if we fix a $\sigma$-algebra, and define a $\sigma$-additive finite measure on it, we can integrate any bounded measurable functions (which is such a nice thing after I struggled on my first year with Riemann integrability). However, yet again - is it just a nice, neat and beautiful way of doing this, or is it the way and this is way the requirements on the measurability of maps are not just due to the (relative) simplicity of dealing with them.

Perhaps, there is also a possible reason for the need of measurability besides the information structure modeling or integrating nice functions, but I don't know what could it be. I must say that I love measure-theory a lot, and found it extremely beautiful - I just want to understand myself (and also to be able to convince others) why shall we deal with measurable structures.

Summary: in a view of the answer by Alexander, comments of Yuri and a conversation I had with Michael Greinecker, being asked the entitling question, I would argue that indeed on can define integration and probability without the notion of measurability - but in such a case one may loose many intuitive properties (e.g. LLN) which would make the interpretation of the result harder. In a particular case of the optimal control, I guess the argument could be: applying a non-measurable strategy is permitted, yet may no be able to which value would it yield, and what does this value mean (e.g. without LLN).

-
Even earlier, Bruno De Finetti (en.wikipedia.org/wiki/Bruno_de_Finetti) also tried to settle the Theory of Probability without assuming countable additivity, with deep philosophical arguments. But after all, mathematically, the study of additive probabilities may be brought back to sigma-additivity. – Pietro Majer Jul 16 '13 at 14:23
@PietroMajer: indeed, and his research is also cited in the section in Dubins and Savage that I've mentioned in OP. So do you mean that the requirement of measurability is mostly a technical drawback of a $\sigma$-additive setting? – Ilya Jul 16 '13 at 14:33
Well, already the case of the Lebesgue measure shows that in general one can't hope to have a sigma additive measure defined on the whole power set. Once one agrees to have measures defined on sigma algebras, I'd say measurability of maps is indeed a must. – Pietro Majer Jul 16 '13 at 15:04
Pietro and Benoit, you are both incorrect. It is consistent with ZFC that there are sigma additive probability measures defined on the whole power set, even extending Lebesgue measure on R. See "real valued measurable cardinals." – Monroe Eskew Jul 16 '13 at 20:08
Benoit, When dealing with abstract probability theory, why is the Lebesgue measure so important? Probability spaces need not be geometrical. – Monroe Eskew Jul 18 '13 at 0:54

You can do many more things with measurable functions than with nonmeasurable ones, so the theory is richer and has more applications. There's much less trouble with integrals, for example, right? In fact, if in the beginning the measurability seems a necessary evil, in the end it becomes a blessing and helps interpreting things. E.g., if a random variable $Y$ is measurable with respect to a sigma-algebra generated by a random variable $X$ then $Y$ can be expressed in terms of $X$, i.e., $Y=f(X)$ for some $f$. Similarly, many concepts of dependence like the Markov property, martingale property, notions from stochastic control, etc. are natural to formulate in terms of measurability w.r.t. appropriately chosen sigma-algebras.

-
Thanks for the answer, and +1. Actually, I was going to include the digression on measurability as an information structure or as a regularity structure in the OP, but I thought that it would overwhelm it. In fact, for some $f$ is in fact for some measurable $f$, and the existence of non necessarily measurable $f$ just requires level sets of $Y$ to be saturated w.r.t. those of $X$. Similarly, the information structure can be defined just based on the partitions rather than $\sigma$-algebras, which are similar in nature - but don't require any regularity which appears to be just additional – Ilya Jul 17 '13 at 8:23
...due to this reason, I would say that measurability is a way to model the dependence concepts, but it is not the way of doing it. Provided integrals do make sense, Markov property, martingales and sequential control concepts could be defined with invoking an "expensive" notion of measurability. Being able to integrate is perhaps a stronger argument, and so far I do not see any other actual reason for the measurability. However, yet again - maybe integration comes easier with measurability, but the latter is still not the must? – Ilya Jul 17 '13 at 8:29
Partitions are certainly very useful. Isn't a partition required to agree with a sigma-algebra though? – Yuri Bakhtin Jul 17 '13 at 14:31
I meant that e.g. in game theory to model the information, one sometimes uses partitions of the space of histories without any mentioning of $\sigma$-algebras. A function on a history space is called measurable w.r.t. partition just if its level sets are saturated w.r.t. partition sets. So of course there would be a $\sigma$-algebra generated by such partition, but a function measurable w.r.t. partition (=its values are constant on partition sets) may be not regular enough to be measurable w.r.t. some $\sigma$-algebras that contains such partition. – Ilya Jul 17 '13 at 14:35
... as an example, every function is measurable w.r.t. partition that separates all points of, say $[0,1]$. On the other hand, for any $\sigma$-algebra different from $2^{[0,1]}$ there is a non-measurable set $A$, so $1_A$ would be a non-measurable function w.r.t. $\sigma$-algebra. – Ilya Jul 17 '13 at 14:37

Regarding measurability: Depending on the audience, it may help to say that standard limit theorems (LLN, CLT, etc.) require measurability (here's something very simple I wrote recently on what you can say in the way of LLN without measurability--basically nothing, except for what you get by applying the standard W/S LLN to the measurable minorant/majorants). Most of the practical applications of probability theory depend in some way on limit theorems: e.g., statistics and CLT, stochastic algorithm convergence, Bayesian convergence, etc.

Regarding finite additivity: On the bright side, if one replaces countable with finite additivity one gets to keep some limit theorems (at least those that give explicit bounds on convergence of distributions: just approximate r.v.'s on a finitely additive space with simple ones, and apply the standard theorems to the simple ones).

On the dark side, replacing countable additivity with finite additivity will not make all sets measurable in dimensions $n\ge 3$, due to Banach-Tarski issues. In situations where there are rotational symmetries, one will still have to have non-measurable sets, assuming AC.

-
Very nice, and thanks. Indeed loosing limit theorems is a strong argument. Regarding finite additivity: from what little I've seen, f.a. measures were initially defined over algebras (thus having a domain of definition somewhat smaller that c.a. measures), and still were related to the measurability. At the same time, the book I mentioned seemed to use f.a. measures defined over the powerset. Is the true that any f.a. measure can be extended from the algebra it is initially defined over to the powerset? I guess, your comment on Banach-Tarski says the answer is "no", am I right? – Ilya Jul 29 '13 at 8:10
Banach-Tarski shows that you can't always extend a f.a. measure to the powerset while keeping isometric invariance on $\mathbb R^n$ ($n\ge 3$). But if you don't require isometric invariance, you can have finitely additive measures over the whole powerset. In fact, given AC, you can have a translation (but not rotation) invariant f.a. measure on all of $2^{\mathbb R^n}$, normalizing the unit cube: the translation group is Abelian, hence amenable, and so we can apply Tarski's Theorem. – Alexander Pruss Aug 6 '13 at 15:26
The answer to your extension question seems affirmative. (1) Any f.a. measure $\mu$ defined on a proper subalgebra $F$ of the powerset extends somewhat. Take a subset $A$ not in $F$. Any set in the algebra gen. by $A$ and $F$ is of the form $(A\cap B)\cup (A^c\cap C)$ for $B,C\in F$. Let the measure of that subset be $\mu(C)$ (exercise: well-defined). So we have an extension from $F$ to the algebra generated by $F$ and $A$. (2) Define an ordering on all f.a. measures by $\mu\le\nu$ iff $\nu$ extends $\mu$. By Zorn (exercise), there is a maximal measure. By (1), it is defined on the powerset. – Alexander Pruss Aug 6 '13 at 15:38
You also want countable additivity to avoid the problems of nonconglomerability. – Alexander Pruss Aug 6 '13 at 17:46
Interesting, let me try work out this procedure - I guess if exists the maximal measures will not be unique. I never heard of conglomerability before - shall I look in the work by de Finetti for this concept? – Ilya Aug 7 '13 at 6:43