# Finding an example of exterior measure

I am finding a sequence of set $$E_1\supset E_2 \supset ...$$,and $$m^*(E_k)<\infty$$,satisfy$$m^*(\cap_{k=1}^{k=\infty}E_k)<\lim_{k\rightarrow \infty}m^*E_k$$

Attempt I think maybe I should use the unmeasurable set .because if a sequence of set named $$A_n$$ is measurable then $$m^*(\cap_{k=1}^{k=\infty}E_k)=\lim_{k\rightarrow \infty}m^*E_k$$

But I can’t construct the sequence of $$E_n$$, can someone help me thanks a lot

Definition $$m^*(E)=inf\{mG|E\subset G,\text{G is an open set}\}$$

• I have deleted my careless answer. – Jochen Apr 16 '19 at 14:02
• @Jochen never mind ~ – jackson Apr 16 '19 at 14:03
• Which outer measure do you consider? $m^*(E)=0$ for $E\subseteq \mathbb R$ finite and $m^*(E)=1$ for $E$ infinite defines an outer measure where you can easily find examples, e.g., $E_n=[n,\infty)$. – Jochen Apr 16 '19 at 14:08
• @Jochen I am a new learner of real analysis ,I edited my question – jackson Apr 16 '19 at 14:12
• For the Lebesgue outer measure on $\mathbb R$ this is more difficult. As far as I remember in the book Measure and Category of Oxtoby you can find (or use some result there to show existence) of a family of disjoint subsetes $A_x$ (with $x\in\mathbb R$) of $[0,1]$ each having outer Lebesgue measure $1$ . Then you can take $E_n=\bigcup\limits_{k\in \{n,n+1,\ldots\}} A_k$. – Jochen Apr 16 '19 at 14:14

EDIT. I think that the standard example of a non-measurable subset of $$\mathbb R$$ can be used: Let $$A\subseteq [0,1]$$ contain exactly one element of each equivalence class of the relation $$x\sim y$$ if $$x-y\in\mathbb Q$$. Then $$m^*(A)$$ is finite and strictly positive because of $$\mathbb R=\bigcup_{q\in\mathbb Q} (A+q)$$ and the sub-$$\sigma$$-additivity of the outer Lebesgue measure. Moreover, the sets $$q+A$$ are pairwise disjoint. Now take $$E_n=\bigcup\limits_{k\in \{n,n+1,\ldots\}} (A+1/k)$$. The intersection is empty but the outer measure of each $$E_n$$ satisfies $$m^*(A)\le m^*(E_n)\le 2$$.
• In my question $m^*(E_k)<\infty$ – jackson Apr 16 '19 at 13:53
• Thanks a lot !and I think in your answer q+E means $q+E_n.$ – jackson Apr 17 '19 at 8:58
• but what about lim$m^*E_n$ ,is the limit exist? – jackson Jun 18 '19 at 11:34