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

  • $\begingroup$ I have deleted my careless answer. $\endgroup$ – Jochen Apr 16 '19 at 14:02
  • $\begingroup$ @Jochen never mind ~ $\endgroup$ – jackson Apr 16 '19 at 14:03
  • $\begingroup$ 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., $E_n=[n,\infty)$. $\endgroup$ – Jochen Apr 16 '19 at 14:08
  • $\begingroup$ @Jochen I am a new learner of real analysis ,I edited my question $\endgroup$ – jackson Apr 16 '19 at 14:12
  • $\begingroup$ 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 $E_n=\bigcup\limits_{k\in \{n,n+1,\ldots\}} A_k$. $\endgroup$ – 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 $m^*(A)\le m^*(E_n)\le 2$.

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  • $\begingroup$ In my question $m^*(E_k)<\infty$ $\endgroup$ – jackson Apr 16 '19 at 13:53
  • $\begingroup$ Thanks a lot !and I think in your answer q+E means $q+E_n.$ $\endgroup$ – jackson Apr 17 '19 at 8:58
  • $\begingroup$ but what about lim$m^*E_n$ ,is the limit exist? $\endgroup$ – jackson Jun 18 '19 at 11:34

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