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Let $\bigcap_{n=1}^\infty E_n=∅$ and if $\mu^*(E_n) <\infty$ and $E_{n+1} \subseteq E_n $ then $\lim\limits_{n\mapsto \infty} \mu^*(E_n) =0 $ even if each $E_n$ is a non-measurable set, where $\mu^*$ is outer measure. Proof sketch please?

Michael Hardy
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ibnAbu
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  • Of course not. You could have $E_n=[0,1]$ for every $n$. Maybe you meant to assume $E=\emptyset$? Then I'm pretty sure the answer is still no, but the counterexample is not quite so trivial. – David C. Ullrich Apr 29 '16 at 22:46
  • @David thanks for your comment but sorry I just corrected a typo in the question. Please could you read the question again? – ibnAbu Apr 29 '16 at 22:50
  • Really? Something to think about then.... – David C. Ullrich Apr 29 '16 at 22:55
  • Yes it is Theorem 1.15 in the book (left as an exercise to the reader) – ibnAbu Apr 29 '16 at 22:57
  • Interestingly it was used to prove Fundamental Theorem of Calculus for the Lebesgue integral. when the derivative of the function exists except on a countable set. – ibnAbu Apr 29 '16 at 23:00
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    That cannot be true. It is possible to partition $[0,1]$ into a continuum of disjoint sets with outer measure $1$ (see here), so one can partition $[0,1]$ also into a sequence $(F_n)$ of disjoints non-measurable subsets of $[0,1]$ with outer measure $1$ each. Letting $E_n=[0,1]\backslash\bigcup_{i=1}^nF_n$ gives you a counterexample. – Michael Greinecker Apr 29 '16 at 23:18
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    No, it's false. – David C. Ullrich Apr 29 '16 at 23:24
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    I actually looked at Theorem 1.15 in that book, it says something entirely different. – Michael Greinecker Apr 29 '16 at 23:24
  • @Michael. Please could you prove theorem 1.15? – ibnAbu Apr 30 '16 at 02:43
  • @Michael I notice a problem in your disproof: intersection of all your $ E_n $ is not necessarily a null set. I observed the construction of the non measurable set based on axiom of choice was based on the fact outer measure is translational invariant. – ibnAbu Apr 30 '16 at 03:46
  • @MichaelGreinecker : Proper usage is $E_n=[0,1] \setminus \bigcup_{i=1}^n F_n$, not $E_n=[0,1]\backslash\bigcup_{i=1}^nF_n$. $\qquad$ – Michael Hardy Apr 30 '16 at 06:32
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    @stalker2133 That $(F_n)$ is a Partition means that every point in $[0,1]$ lies in some of the $F_n$ and is therefore not in $E_n$. Since $\bigcap_n E_n$ is the set of points that are in every $E_n$, this intersection is empty and contains no point at all. If you make such statements, please prove them. You haven't given a counterexample an there is none. – Michael Greinecker Apr 30 '16 at 11:06
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    @MichaelHardy ??? I don't see the difference between what you say is the right version and the wrong version. There was an obvious typo (not a problem with "proper usage", a slip-of-the-finger error): He obviously meant $E_n=[0,1]\setminus\bigcup_{i=1}^nF_i$. – David C. Ullrich Apr 30 '16 at 12:10
  • @MichaelGreinecker Too bad we can't edit old comments - you might want to clarify the typo: You wrote $\bigcup E_n$ for the intersection... – David C. Ullrich Apr 30 '16 at 12:35
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    @DavidC.Ullrich Thanks for the comment,it should be correct now. – Michael Greinecker Apr 30 '16 at 16:13
  • @MichaelGreinecker Heh. I can't edit comments after five minutes - what do I have to do to get magic powers like yours? – David C. Ullrich Apr 30 '16 at 17:08
  • @MichaelGreinecker : Using \setminus results in spacing appropriate to a binary operator or binary relation, and \backslash does not. $\qquad$ – Michael Hardy Apr 30 '16 at 17:47
  • @DavidC.Ullrich Get elected. – Michael Greinecker Apr 30 '16 at 22:24

1 Answers1

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It's false. (I was taking a shower when Michael posted his comment; what's below is a detailed exposition of a simpler version of what he said.)

Say $G$ is the group $[0,1)$, with addition modulo $1$. Note that Lebesgue outer measure is $G$-invariant. I'll be writing $a+b$ for the addition in $G$.

Let $H=[0,1)\cap\Bbb Q$, and let $C$ be a complete set of coset representatives for $H$ as a subgroup of $G$. Note that every $x\in G$ has a unique representation as $h+c$ with $h\in H$, $c\in C$.

Let $H=\{h_1,h_2,\dots\}$, and then define $H_n=\{h_n,h_{n+1},\dots\}$. Define $$E_n=H_n+C.$$

Then $\cap E_n=\emptyset$, although for every $n$ we have $$\mu^*(E_n)\ge\mu^*(h_n+C)=\mu^*(C)>0.$$

David C. Ullrich
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  • Check carefully the intersection of all your $ E_n $ is not necessarily a null set. – ibnAbu Apr 30 '16 at 03:48
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    @stalker2133 The intersection of the $E_n$ is empty. Suppose $x\in G$. Then $x$ has a unique representation $x=h+c$ with $h\in H$ and $c\in C$. There exists $n$ so $h=h_n$. Hence if $k>n$ then $x\notin E_k$, since $h_n\notin H_k$. – David C. Ullrich Apr 30 '16 at 12:06
  • @stalker2133 Trying to imagine what you might be misunderstanding here, maybe it's the notation $A+B$. If $A$ and $B$ are subsets of an abelian group then $A+B={a+b:a\in A,b\in B}$. – David C. Ullrich Apr 30 '16 at 12:34
  • I Agreed with the counter example. Problem closed. – ibnAbu May 02 '16 at 16:01