0

I have seen a few of other questions on here that provide examples of sets that are Lebesgue measurable but not Borel measurable, but I believe I have a lapse in understanding somewhere. I know that the Lebesgue measure is a special case of a Borel measure, so then are they not both defined on the sigma algebra generated by all Borel sets?

CBBAM
  • 7,149
  • 3
    Lebesgue measure is defined on a strictly larger sigma algebra, on which every subset of a set of measure zero is measurable, i.e. Lebesgue measure is complete. This property does not hold for Borel measure. –  Dec 06 '20 at 00:07
  • @Bungo So the Lebesgue measure is just the completion of of any Borel measure? That is, the domain only differs in that it contains all subsets of nullsets (null with respect to Lebesgue)? – CBBAM Dec 06 '20 at 00:08
  • 1
    Yes, Lebesgue measure is the completion of Borel measure. The domain of Lebesgue measure differs in that it is the smallest sigma-algebra that contains the Borel sets as well as all subsets of nullsets. –  Dec 06 '20 at 00:10
  • 1
    A precise statement is $A$ is Lebesgue measurable if and only if $A =B \cup C$ where $B$ is a Borel set and $C$ is a subset of some Borel set with measure $0$. – Kavi Rama Murthy Dec 06 '20 at 00:11
  • @Bungo Thank you, just one more question. Given that Borel sets are generated by h-intervals, closed intervals, or open intervals, would it by default not contain any possible set in $\mathbb{R}$? Maybe this is where my intuition is leading me astray... – CBBAM Dec 06 '20 at 00:12
  • @KaviRamaMurthy So $A$ in that case isn't in the Borel sigma algebra? Isn't it a finite union of sets that are also Borel? – CBBAM Dec 06 '20 at 00:12
  • 1
    No, there are subsets of $\mathbb R$ that are not Lebesgue measurable, hence also not Borel measurable. To construct such a subset you need the axiom of choice. For details you can see the Vitali set wikipedia page. –  Dec 06 '20 at 00:13
  • @Bungo Thank you, I will take a look. – CBBAM Dec 06 '20 at 00:15
  • 2
    For an example of a set that is Lebesgue measurable but not Borel measurable, you can see this MSE answer which gives a standard construction. Essentially you start with a non-Lebesgue measurable set such as the Vitali set (as in my previous comment) and "compress it" so that it fits inside the Cantor set, which has measure zero (and is both Borel and Lebesgue measurable). –  Dec 06 '20 at 00:16
  • 1
    In my previous comment $C$ need not be a Borel set. It is a subset of a Borel set of measure $0$. – Kavi Rama Murthy Dec 06 '20 at 00:20
  • @CBBAM Why not go through th Caratheodory construction (via outer measure) of the Lebesgue measure? Then you can see the relation between the sigma algebra of Lebesgue measurable sets and the Borel sigma algebra. – Matematleta Dec 06 '20 at 00:45
  • Based on some of the recent Q's here, I have the impression that there is a new style of teaching measure theory without mentioning inner measure. Let $m$ be Lebesgue measure. Let $m^o$ be Lebesgue outer measure. The Lebesgue inner measure $m^i(S)$ is $\sup {m(T): T\in C_S}$ where $C_S$ is the set of compact subsets of $S$. If $S$ is a bounded subset of $\Bbb R$ then $S\in dom(m)$ iff $m^o(S)=m^i(S).$ And if $S\in dom(m)$ then $m(S)=m^o(S)=m^i(S)$..... (continued) .... – DanielWainfleet Dec 06 '20 at 05:07
  • ...(continued).... Using some set theory we can show ("abstractly") that there exists a Bernstein set $B.$ https://en.wikipedia.org/wiki/ . Every member of $C_B$ and every member of $C_{\Bbb R\setminus B}$ is countable, so $m^i(B)=m^i(\Bbb R\setminus B)=0.$ It follows that $B\not\in dom(m).$ – DanielWainfleet Dec 06 '20 at 05:25

0 Answers0