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I have read that probability measure cannot be defined on set of all subsets of unit interval namely (0,1]. Proof uses construction of the Vitali set etc. Specifically, it is well known that probability measure with the following properties does not exist on power set of (0,1].

  1. Translational invariance
  2. If $0 \leq a \leq b \leq 1$, then $\mathbb{P}((a,b]) = b-a$

This is actually Lebesgue measure on (0,1].

This might be a dumb question but I cannot find an answer to it. The domain of probability measure is a Borel Sigma algebra while that of a Lebesgue measure is a Lebesgue Sigma algebra. It is well known that Lebesgue sigma algebra has cardinality of $2^{\mathbb{R}}$ while that of Borel sigma algebra is $2^{\mathbb{N}}$.

So even though Lebesgue measure on (0,1] $ \textbf{IS}$ a probability measure satisfying 1. and 2. above, its domain has a cardinality strictly greater than that of Borel Sigma Algebra.

Where am I going wrong? Even though Lebesgue measure on (0,1] is in fact a uniform probability measure, why is there a difference in the domains? If not, please clarify.

TryingHardToBecomeAGoodPrSlvr
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  • A Lebesgue measurable set can be written as a union of a Borel set and a set of measure zero (which may not be Borel measurable). – Eclipse Sun Feb 03 '19 at 21:45
  • Statement A: Uniform probability measure with translation invariance cannot be defined on power set of (0,1]. Statement B: Borel measure is same as Lebesgue measure on (0,1]. Statement C: Lebesgue Sigma Algebra has a cardinality strictly greater than that of Borel Sigma Algebra. Now statement B must imply that domains of Borel and Lebesgue measures are same in (0,1]. But then, statement C is implying that they have to be different. Where am I going wrong? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 21:50
  • What do you mean by "uniform probability measure"? B is wrong and C is correct. – Eclipse Sun Feb 03 '19 at 22:09
  • Uniform probability measure is defined as one which has the translational invariance property. Why is B wrong? Basically, the measure is b-a on sets (a,b] which are in (0,1] and they are both sigma algebras containing all the intervals in (0,1]. They are exactly the same right? They will differ when you define them on all of $\mathbb{R}$. – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 22:13
  • @TryingHardToBecomeAGoodPrSlvr Why does "same on intervals" imply "same on every subset"? – Clement C. Feb 03 '19 at 22:16
  • @TryingHardToBecomeAGoodPrSlvr Then Statement A is wrong because Lebesgue measure and Borel measure are both translation-invariant. As I wrote in the first comment, the Lebesgue $\sigma$-algebra is essentially the Borel $\sigma$-algebra adding the null sets. – Eclipse Sun Feb 03 '19 at 22:18
  • The Borel $\sigma$-algebra is a strict subset of the Lebesgue $\sigma$-algebra, so your statement B is not strictly correct. However, if you interpret it loosely to mean no more than that Borel measure is the restriction of Lebesgue measure to the Borel sets, then it would be correct. The Lebesgue $\sigma$-algebra is just the union of the Borel $\sigma$-algebra with the collection of all subsets of any Borel sets that have measure $0$. – lonza leggiera Feb 03 '19 at 22:19
  • Motivation for me to ask this question was the first line in my main post. We have to "settle" to a smaller sigma algebra like a Borel Sigma Algebra because uniform probability measure (ie with properties 1. and 2. in the main post) cannot be defined on power set of real numbers. Now you are saying that Lebesgue measure is a uniform probability measure. It is also well known that Lebesgue Sigma Algebra is equicardinal with that of power set of real numbers. So then why even bother to restrict probability measures to Borel Sigma Algebra? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 22:25
  • With some kind of bijection on Lebesgue sigma algebra and power set of (0,1] (which must be possible as they are equicardinal), we must be able to define a uniform probability measure on power set of (0,1]. But then there is a proof that such an assignment is not possible. So where am I going wrong in this chain of thought? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 22:26
  • I found this link which says that Satement B is correct. https://math.stackexchange.com/questions/2657706/lebesgue-measure-restricted-to-a-set-of-measure-1-is-it-a-probability-measure – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 22:42
  • @TryingHardToBecomeAGoodPrSlvr How is the link you gave saying that? (sorry if I am missing something, but it just states that the Lebesgue measure on $[0,1]$ is a probability measure. This is not what you call "B") – Clement C. Feb 03 '19 at 22:59
  • In that link, they essentially say that the probability measure is also uniform which implies translational invariance. – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:00
  • But how does that mean it's the same as the Borel measure on $[0,1]$ (which is what you call "B")? Some sets are measurable under one and not the other... @TryingHardToBecomeAGoodPrSlvr – Clement C. Feb 03 '19 at 23:03
  • Point taken. Some sets are not Borel measurable when the domain is Lebesgue sigma algebra. But it nevertheless is a uniform probability measure as confirmed in the link above. The point is the following. We have a uniform probability measure on (0,1]. Domain is Lebesgue sigma algebra which is equicardinal with power set of (0,1]. So we can have a uniform probability measure defined on power set of (0,1]. This is in contradiction with the statement "there are no uniform probability measures on the power set of (0,1]." – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:11
  • The proof uses Vitali set and is a well known proof. So where is the mistake in this chain of thought? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:13
  • Borel Lebesgue measure and the full Lebesgue measure are by definition not the same, the latter extends the former to more subsets. But there is no realistic prospect of defining it beyond Lebesgue measurable sets. – Henno Brandsma Feb 03 '19 at 23:36
  • https://math.stackexchange.com/questions/1769084/prove-that-lebesgue-measurable-set-is-the-union-of-a-borel-measurable-set-and-a – tryst with freedom Apr 03 '23 at 22:00

1 Answers1

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We have the standard Lebesgue measure $\mu$, that is defined on all Borel sets, and obeys translation invariance and interval-consistency (my name for $\mu((a,b])=b-a$ for all relevant intervals; you call it uniform in the comments) . Indeed there are $\mathfrak{c}$ many Borel sets on which this $\mu$ is then defined.

Independently of that we can define null sets as:

$A \in \mathscr{N}$ iff for every $\varepsilon >0$ we can find at most countably many open intervals $(a_n, b_n)$ such that $A \subseteq \bigcup (a_n, b_n)$ and $\sum_n |b_n - a_n| < \varepsilon$ and check that such subsets are closed under subsets and countable unions (they form a $\sigma$-ideal). There are $2^\mathfrak{c}$ many null sets (e.g. as the standard Cantor set is one and so are all of its subsets.)

It turns out we can extend $\mu$ to a measure on more subsets, namely take the collection of all sets of the form $B \cup N$ where $B$ is Borel, and $N$ is a null set and just define $\mu'(B \cup N) = \mu(B)$. One can prove that this is well-defined (e.g. if a Borel set happens to be a null set, $\mu$ was $0$ on it anyway, justifying the name) and greatly extends the domain of $\mu$, while keeping translation invariance.

The existence of a Vitali set (if we assume AC) shows that we cannot extend $\mu'$ even further to $\mathscr{P}(\mathbb{R})$ while keeping translation invariance intact. $\mu'$ is called the completion of $\mu$ and it's also obtained when we apply the Carathéodory theorem on the Lebesgue measure on the half-open interval algebra; we get (IIRC) the same class of Lebesgue measurable subsets.

It depends on your application whether you want to work with only Borel sets or all Lebesgue measurable sets. Mostly in analysis the latter is done; taking the larger domain.

If we give up translation invariance and so "neutralise" the Vitali set examples (whose proof of non-measurability hinges on this translation invariance property) there are still obstacles: under CH ($\mathfrak{c}=\aleph_1$) there can be no finite measure on $(0,1]$ that measures all subsets and that gives all singletons measure $0$ (much weaker than being uniform), as was shown by Ulam. So you could never define such a measure on all subsets without additional set-theoretic assumptions. See also real valued measurable cardinals etc. It gets complicated.

Henno Brandsma
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  • Thanks for the nice answer. I am still relatively new to measure theory so I could not understand some parts of your answer. The part that caught my attention was the statement "and greatly extends the domain of , while keeping translation invariance." Does that imply that Lebesgue measure, with the domain being the Lebesgue $\sigma - algebra$, is a probability measure with the property of translational invariance? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:36
  • @TryingHardToBecomeAGoodPrSlvr Yes, this is clear from the definition I gave as well. Null sets are also translation invariant. – Henno Brandsma Feb 03 '19 at 23:37
  • I thought that it was not possible to have translational invariant probability measures defined on sets with cardinality $2^\mathbb{R}$ whose proof can be based off of something like that given in https://math.stackexchange.com/questions/1544458/proving-that-theres-no-translation-invariant-measure-on-the-power-set-of-math. Not sure where I am going wrong here. – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:38
  • @TryingHardToBecomeAGoodPrSlvr You cannot define it on all subsets of $\Bbb R$, but the Vitali set is not Lebesgue measurable. The size of the $\sigma$-algebra has no role here, there are as many Lebesgue measurable (even null) sets as there are subsets of the reals, but that doesn't allow us to extend a measure in a sensible way.. – Henno Brandsma Feb 03 '19 at 23:41
  • Awesome! So it is possible to have a translationally invariant uniform probability measure on set of all subsets of (0,1] but not on all subsets of $\mathbb{R}$ right? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:45
  • @TryingHardToBecomeAGoodPrSlvr No, on neither. Just on Lebesgue measurable subsets! – Henno Brandsma Feb 03 '19 at 23:46
  • OK so not on power set of $\mathbb{R}$ but on the Lebesgue sigma algebra of (0,1]. This is where it gets confusing. If I could define a translationally invariant probability measure on Lebesgue sigma algebra of (0,1] (call it L), whose cardinality is same as that of power set of (0,1] (which is $2^\mathfrak{c}$), why is it not possible to extend it to the power set of (0,1]? The reason I am claiming that is because there is a bijection between Lebesgue sigma algebra and power set of (0,1]. – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:55
  • So P mapping from L to [0,1] can be split to P mapping from $2^\mathfrak{c}$ to L to [0,1], the middle being possible as they are equicardinal. The only problem might be that translational invariance may not hold anymore. Is that the case? – TryingHardToBecomeAGoodPrSlvr Feb 03 '19 at 23:55
  • Let us continue this discussion in chat. – TryingHardToBecomeAGoodPrSlvr Feb 04 '19 at 00:32