# General property regarding outer measure for a nested sequence of sets (measurable or not).

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?

• Actually it is true and it was given as an exercise in 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 )' by Russell A. Gordon – ibnAbu Apr 29 '16 at 22:53
• 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
• No, it's false. – David C. Ullrich Apr 29 '16 at 23:24
• I actually looked at Theorem 1.15 in that book, it says something entirely different. – Michael Greinecker Apr 29 '16 at 23:24
• @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

## 1 Answer

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 $$\mu^*(E_n)\ge\mu^*(h_n+C)=\mu^*(C)>0.$$

• Check carefully the intersection of all your $E_n$ is not necessarily a null set. – ibnAbu Apr 30 '16 at 3:48
• @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 2 '16 at 16:01
• I gave a proof of theorem 1.15 stated in the book math.stackexchange.com/questions/1772057/… – ibnAbu May 7 '16 at 18:44