Timeline for Why do we want maps to be measurable (in countably-additive setting)
Current License: CC BY-SA 3.0
12 events
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| Jul 19, 2013 at 9:05 | vote | accept | SBF | ||
| Jul 17, 2013 at 15:56 | comment | added | SBF | I see - I thought you meant the existence of conditional expectation. | |
| Jul 17, 2013 at 15:38 | comment | added | Yuri Bakhtin | These are not really results but definitions. Those of conditional expectations w.r.t. sigma-algebras or r.v.'s. | |
| Jul 17, 2013 at 15:03 | comment | added | SBF | Your statement there are less meaningful things you can say about them makes perfect sense to me, and perhaps now that's what I would use a reply to my hypothetical questioner - even in case this a similar attitude as people previously had to smooth functions. Could you include this in your answer so that I will accept it? I'm not sure, thought, that I got your comment on Kolmogorov's work - do you mean that these results required measurability? | |
| Jul 17, 2013 at 14:53 | comment | added | Yuri Bakhtin | Nobody forbids using those nonmeasurable functions, it's just there are less meaningful things you can say about them. Well, this sounds a little like the attitude towards nonsmooth functions a century or two ago, so you have all rights to question this (although I suggest worrying about more contentful stuff instead). Anyway, here's one important point that is often overlooked: in Kolmogorov's book on foundations of probability not only he introduced measure theory to probability, but also the rigorous concepts of dependence and conditional expectation and conditional probability. | |
| Jul 17, 2013 at 14:37 | comment | added | SBF | ... 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. | |
| Jul 17, 2013 at 14:35 | comment | added | SBF | 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. | |
| Jul 17, 2013 at 14:31 | comment | added | Yuri Bakhtin | Partitions are certainly very useful. Isn't a partition required to agree with a sigma-algebra though? | |
| Jul 17, 2013 at 8:29 | comment | added | SBF | ...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? | |
| Jul 17, 2013 at 8:23 | comment | added | SBF | 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 | |
| Jul 16, 2013 at 20:27 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Jul 16, 2013 at 19:05 | history | answered | Yuri Bakhtin | CC BY-SA 3.0 |