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](http://web.mit.edu/dimitrib/www/soc.html), or a [gambling](http://books.google.nl/books/about/Discrete_Gambling_and_Stochastic_Games.html?id=78W2z0SeVDYC&redir_esc=y) 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](http://books.google.nl/books/about/How_to_gamble_if_you_must.html?id=kt9QAAAAMAAJ&redir_esc=y) 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).



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


