Timeline for Why do we need sigma-algebras to define probability spaces?
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| Mar 16, 2022 at 14:50 | comment | added | user266286 | I didn't go through the whole post, but what I read is very well written. Minor quibble: You said ''A theory of probability admitting all subsets of uncountable sets will break mathematics'; not quite. If $X$ is uncountable and $x\in X$ then we can trivially define a probability function on all subsets of $X$: let $P(A)=1$ if $x$ is in $A$ and let it be $0$ otherwise. No useful definition of probability is possible however; this is just a pathological case. | |
| Jun 29, 2020 at 2:48 | comment | added | Sycorax♦ | @Iamanon You’d have to ask the instructors of those courses to get a fair treatment. | |
| Jun 28, 2020 at 18:38 | comment | added | 24n8 | If $\sigma$-algebra fixes math and is so useful, why do probability courses (even the ones at the graduate level) rarely introduce this concept? | |
| Jun 11, 2020 at 14:32 | history | edited | CommunityBot |
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| Jan 19, 2020 at 18:27 | history | edited | Alexis | CC BY-SA 4.0 |
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| Mar 13, 2019 at 2:32 | comment | added | Sycorax♦ | @YatharthAgarwal Looks good to me! +1 | |
| Mar 13, 2019 at 2:22 | comment | added | Yatharth Agarwal | @Sycorax Ah, thank you for the quick responses. I was going crazy all day. I asked the question here. Do you think it’s well-asked / gets at the heart of the point? | |
| Mar 13, 2019 at 1:26 | comment | added | Sycorax♦ | @YatharthAgarwal Ah! I understand. I see now how there's a logical gap in my answer where I go to describing Lebesgue measure. I don't have an example at hand, but I think it would be a good question to ask. | |
| Mar 13, 2019 at 0:21 | comment | added | Yatharth Agarwal | @Sycorax Antoni’s example justifies why we need the structure of event spaces, $\sigma$-algebras, and measure theory. It does not, however, provide a concrete example of a paradox that arises from, say, taking $\mathcal F = 2^\Omega$ for non-trivial uncountable $\Omega$. | |
| Mar 13, 2019 at 0:08 | comment | added | Sycorax♦ | @YatharthAgarwal I believe Antoni's example in the original post provides an example that fits this description. | |
| Mar 12, 2019 at 23:25 | comment | added | Yatharth Agarwal | @Sycorax You talk about Vitali’s set and Banach-Tarski and Lebesgue measures. But probability spaces do not require translation-invariant measures, rotation-invariant measures, or Lebesgue measures. Would you have a different example of a born of uncountable sets that does apply to probability spaces? (Please forgive me if I’ve made an error; I’m new to measure theory!) | |
| Sep 29, 2018 at 22:47 | comment | added | Theo | I really enjoyed reading your answer. I don't know how to thank you, but you've clarified things a lot! I've never studied real analysis nor had a proper introduction to mathematics. Came from an Electrical Engineering background that focused a lot on practical implementation. You've written that in so simple terms that a bloke like me could understand it. I really appreciate your answer and the simplicity you've provided. Also thanks to @Xi'an for his packed comments! | |
| Sep 14, 2018 at 1:55 | history | edited | Sycorax♦ | CC BY-SA 4.0 |
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| Aug 10, 2017 at 15:35 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Jul 30, 2017 at 21:46 | comment | added | hyiltiz | Do you mean I ask as a separate question? I think it would make this answer more complete, so I am reluctant to ask a separate one. | |
| Jul 30, 2017 at 18:24 | comment | added | Sycorax♦ | @hyiltiz I think that question would be a fine one to ask. | |
| Jul 30, 2017 at 0:56 | comment | added | hyiltiz | Would be even greater if the answer touched upon conditional probability in the same style. | |
| Apr 12, 2017 at 7:06 | history | bounty ended | Tim | ||
| Mar 8, 2016 at 21:44 | comment | added | Elvis | If you drop the axiom of choice, you can add an axiom stating that all subsets of R are Lebesgue measurable, and still have a coherent theory... | |
| Mar 8, 2016 at 21:27 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 8, 2016 at 20:37 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 3, 2016 at 23:57 | vote | accept | Antoni Parellada | ||
| Mar 3, 2016 at 14:30 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 2, 2016 at 12:46 | comment | added | amoeba | @Student001: I think we are splitting hairs here. You are right that the general definition of "measure" (any measure) is given using the concept of sigma-algebras. My point, however, is that there is no word or concept of "sigma-algebra" in the definition of the Lebesgue measure provided in my first link. In other words, one can define Lebesgue measure as per my first link but then one needs to show that it is a measure and that's the hard part. I agree that we should stop this discussion though. | |
| Mar 2, 2016 at 12:45 | comment | added | KOE | I find this discussion interesting, but as I've understood it comments aren't for discussions so I'll try and not clutter further. I'm open to chatting about it with anyone equally interested in the details. | |
| Mar 2, 2016 at 12:38 | comment | added | KOE | @amoeba Maybe we read the first link differently. To me, they seem to follow the usual route. Outer measure first on space $\Omega$, find a $\sigma-$algebra $\mathscr F$, then define the measure on the measurable space $(\Omega, \mathscr F)$. The second link call a set function on a ring a measure. That's interesting and may be were the confusion comes from. I've never seen a measure defined on anything else than a $\sigma-$algebra. In fact, I don't understand what definition of measure that would allow that. | |
| Mar 2, 2016 at 11:34 | comment | added | amoeba | @Student001: What you are saying is correct, but there is nothing circular in defining $\mathscr F$ as the set of all Lebesgue-measurable sets, as user777 did. You don't need the concept of $\sigma$-algebra to define Lebesgue measure, see wiki. Of course when Lebesgue measure is introduced it is not enough just to give this definition; one needs to show that it exists, that it is unique, etc.; the concept of $\sigma$-algebra appears on the way. | |
| Mar 2, 2016 at 3:04 | comment | added | Sycorax♦ | @Student001 Ok, I'll start doing some more reading. My instructor in probability was a measure theorist of all things (who could ask for anything better) so I'll send him a note and see if I can glean some more information, or just a reference. | |
| Mar 2, 2016 at 3:01 | comment | added | KOE | Right! The only way I've seen this defined rigorously is the one outlined at en.wikipedia.org/wiki/Lebesgue_measure where you start with an outer measure, prove that the sets which satisfy a certain condition form a $\sigma-$algebra, and then define the measure as the restriction of the outer measure to this $\sigma-$algebra. That is, the $\sigma-$algebra comes first, then the measure. Your point in the answer is well taken though and it's more of a technical detail than anything else. | |
| Mar 2, 2016 at 2:56 | comment | added | Sycorax♦ | @Student001 Ok, I think I understand what you mean. Taking a shot in the dark at, I think what we've discovered here is that Lebesgue measurability is a byproduct of $\sigma$-fields (or vice-versa), with the point being that measurability is the key point in defining which sets are admitted as events. But without doing more reading, I don't think I can comment any further. Perhaps someone else will chime into this thread (@whuber ...?) | |
| Mar 2, 2016 at 2:43 | comment | added | KOE | You are talking about Lebesgue measure, in this case on the unit square. Then, given the measure $\mathcal L^2$, you define $\mathscr F$ as all the measurable sets, which all are subsets of the unit square in your example. I've never seen a measure defined without it being a function on a $\sigma-$algebra, so I guess I don't understand what $\mathcal L^2$ is here. Let me add that my experience of measure theory is limited to a few classes, so I'm not claiming to be an expert, either. Just interested to know if there's an alternative definition of $\mathcal L^2$. | |
| Mar 2, 2016 at 2:36 | comment | added | Sycorax♦ | @Student001 If I'm reading your comment correctly, then it would seem to say that the Lebesgue measurability of a set is defined as the mapping from a $\sigma$-field to $\mathbb{R}$. I think I'm making the more specific claim about a specific subset of $\mathbb{R}$ and hence which we consider events. But this is just a filtration (a definition, a specification) of the sets under consideration as being events, so I'm not sure if the "direction" of our characterization matters... for the same reason that we can say $0!=1$. | |
| Mar 2, 2016 at 2:32 | comment | added | Sycorax♦ | @AntoniParellada I think that the section discussing the Banach-Tarski paradox is a concrete example. If we can rearrange balls to have arbitrary volume, the same ball can be rearranged however we like and produce a $0=1$ result. | |
| Mar 2, 2016 at 2:30 | comment | added | Sycorax♦ | @Student001 Unless I've been sloppy somewhere, I believe I've been more specific than your characterization. I'm speaking to a specific example in the unit square. Aside from that, I can't say what the usual practice is in measure theory because I have not studied it to any meaningful extent. However, I can say that my (classroom) experience has been that in geometric problems on $\mathbb{R}^n$ is that we define $\mathscr{F}$ as only the $\mathcal{L}^n$ measurable subsets of $\mathbb{R}^n$ to resolve the technical problems I outline. | |
| Mar 2, 2016 at 0:10 | comment | added | Antoni Parellada | It is a great expansion... Coffee is a terrific substance! I'm going to let it play out for a while, among other things, I want to re-read and do some research, but your answer is bound to be a "canonical" post. There is one point that would be beyond good - a concrete example of the contradiction $P(x)=1=0$ you mention. | |
| Mar 2, 2016 at 0:04 | comment | added | KOE | (+1), nice answer! I'm wondering about your example where you define $\mathscr F$ to be the set of all Lebegue-measurable sets, if I read you correctly. I believe I understand the point you are making, but the definition seems circular given that usually we define the mapping $\mathcal L : \mathscr F \to \mathbb R$ | |
| Mar 1, 2016 at 17:15 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 17:09 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 16:58 | comment | added | Sycorax♦ | @Xi'an Thanks for kind words! It really means a lot, coming from you. I was not familiar with the Borel-Kolmogorov paradox as of this writing, but I'll do some reading and see if I can manage to make a useful addition of my findings. | |
| Mar 1, 2016 at 16:54 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 16:46 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 16:18 | comment | added | Xi'an | (+1) Good points! I would also add that without measure and $\sigma$ algebras, conditioning and deriving conditional distributions on uncountable spaces get quite hairy, as shown by the Borel-Kolmogorov paradox. | |
| Mar 1, 2016 at 15:58 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 15:43 | comment | added | Sycorax♦ | @amoeba Referring to my notes, it seems that both are used. | |
| Mar 1, 2016 at 15:41 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 15:39 | comment | added | Antoni Parellada | Thank, @amoeba. I wasn't sure, but I will edit accordingly. | |
| Mar 1, 2016 at 15:38 | comment | added | amoeba | Isn't it much more common to call it a $\sigma$-algebra and not a $\sigma$-field? I have never heard the term "sigma-field" before and the Wiki article is called Sigma-algebra even though it does mention that it is also called "sigma-field". CC to @Antoni who wrote his question using the "sigma field" terminology (even without the hyphen). | |
| Mar 1, 2016 at 15:37 | history | edited | Sycorax♦ | CC BY-SA 3.0 |
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| Mar 1, 2016 at 14:18 | comment | added | Antoni Parellada | Thank you for your answer. $\mathcal{L}$ stands for Lebesque measurable? I'll +1 your answer on faith, but I'd really appreciate it if you could bring down the math level several notches... :-) | |
| Mar 1, 2016 at 14:14 | history | answered | Sycorax♦ | CC BY-SA 3.0 |