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According to this answer the Jordan Measure is not a measure.

This other question points out the fact that the Jordan measurable sets do not form a $\sigma$-algebra.

I am wondering whether the Jordan measure fails to be a measure

  1. because one of the $3$ conditions of a measure is not met

  2. or because there is no $\sigma$-algebra on which it can be defined?

Edit:

If the Jordan measure is a measure, in what $\sigma$-algebra is it defined?

niobium
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  • The set you mention is not Jordan measurable.. (I think). – niobium Oct 05 '22 at 16:08
  • I'd like a proof that for every $\sigma$-algebra the Jordan measure is not a measure – niobium Oct 05 '22 at 16:35
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    Sorry, I got carried away. The issue is that the only sets that can be Jordan measurable are bounded. The Jordan "pseudo-measure" is indeed $\sigma$-additive in the sense that if $(I_n:n\in\mathbb{N})$ are pairwise disjoint J-measurable and $\bigcup_nI_n$ is also J-measurable, then $\mu(\bigcup_nI_n)=\sum_n\mu(I_n)$. This property allows you to extend the psuedomeasure to a measure on a $\sigma$-algebra. – Mittens Oct 05 '22 at 16:40
  • Oh ok, but this $\sigma$-additivity only works for finite coutable unions, right? – niobium Oct 05 '22 at 16:43
  • Ok I get it: as the Jordan measurable sets do not form a $\sigma$-algebra, we have to specify that the countable infinite union is also J-measurable – niobium Oct 05 '22 at 16:53
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    The point is that the way you are constructing the J-measure is by considering the semiring $\mathcal{R}$ of finite union of intervals $(a,b]$ with $-\infty<a\leq b<\infty$ it is not that hard to see that if $I_n$ is a sequence in $\mathcal{R}$ such that $A:=\bigcup_nI_n\in\mathcal{R}$, them $\mu(A)=\sum_n\mu(I_n)$. – Mittens Oct 05 '22 at 17:08
  • So if the Jordan measure has the $\sigma$-additivity property then why do you call it "pseudo-measure"? – niobium Oct 05 '22 at 18:06
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    Because measures are defined on $\sigma$-algebras, and the domain of the Jordan measure is not a $\sigma$-algebra (a ring at best) – Mittens Oct 05 '22 at 18:08

1 Answers1

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Very much late to the party. The Jordan volume/measure/content is defined on the Jordan measurable sets (subsets of $\mathbb{R}^{n}$ whose indicator functions are Riemann integrable). These sets do not form a $\sigma$-algebra because they're not closed under countable unions, as we can see for the classical example $\mathbb{Q}\cap[0,1]$. So, as for reason 1, Jordan volume is not a measure because the indicator function of sets which are countable unions of disjoint Jordan measurable sets may not be Riemann integrable, which means (a) the volume of such unions wouldn't even be defined and (b) Jordan volume is not $\sigma$-additive on Jordan sets. More simply put, is as @Mittens said: not a measure because its domain isn't a $\sigma$-algebra.

As for reason 2, this tells us that the $\sigma$-algebra $\sigma(J)$ $\textit{generated by}$ the collection $J$ of Jordan sets in $\mathbb{R}^{n}$ is formed by the sets which can be written as the union of a borel set and a closed set. So, is Jordan volume a measure in this $\sigma$-algebra? If you keep Jordan volume as the Riemann integral of the indicator function $\chi_{E}$ of a set $E\in\sigma(J),$ you'll still fail because $\chi_{\mathbb{Q}\cap[0,1]}$ is still not Riemann integrable, even though $\mathbb{Q}\cap[0,1]\in\sigma(J).$

Swapping the Riemann integral for the Lebesgue integral and redefining the Jordan volume as $\lambda:\sigma(J)\rightarrow[0,+\infty]$ such that $$\lambda(E)=\int\chi_{E}\ d\mu$$

where $\mu$ is the Lebesgue measure gives you a measure on $(\mathbb{R}^{n},\sigma(J)).$ You can verify this in Bartle's Elements of Integration and Lebesgue Measure, page 31, corollary 4.9. So since without "adapting" the definition of Jordan volume we couldn't make it into a measure, I believe the answer for your question is "both".

Gustavo De Souza
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