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Give an example of a sequence $\{E_i\}$ of disjoint sets such that $m^*(\cup_{i=1}^{\infty}E_i)<\sum_{i=1}^{\infty}m^*(E_i)$.

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  • $\begingroup$ Hi, I guess you mean pairwise-disjoint; just to be a bit more clear? $\endgroup$ – DBFdalwayse Sep 6 '13 at 7:26
  • $\begingroup$ Yes I meant pair wise disjoint. $\endgroup$ – Manjil P. Saikia Sep 6 '13 at 7:37
  • $\begingroup$ Have you seen the construction of a non-measurable set (say, from Royden's book)? $\endgroup$ – Prahlad Vaidyanathan Sep 6 '13 at 7:40
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According to this post here: Vitali set of outer-measure exactly $1$., you can construct a Vitali set for $[0,1]$ with Lebesgue outer measure equal to $1$. On the other hand the Lebesgue outer measure coincide with Lebesgue measure for $[0,1]$ so: $m^*([0,1])=1$. Moreover this interval can be written as the union of disjoint Vitali sets (Vital set and its rational translation). Let $r_i$ be the enumeration of rationals and let $V_i$ be the rational translation of Vitali set by $r_i$. So we have $\displaystyle [0,1]=\bigcup_{i=1}^{\infty} V_i$. But then $$ \displaystyle 1=m^*(\bigcup_{i=1}^{\infty}V_i)<\sum_{i=1}^{\infty}m^*(V_i)=\sum_{i=1}^{\infty}1\rightarrow \infty $$

Another way, is to use usual Vitali set construction over $[0,1]$. Now again set $r_i$, the enumeration of rationals, and $V_i$, the rational translation of Vitali set by $r_i$. We have $\displaystyle [0,1]=\bigcup_{i=1}^{\infty} V_i$ therefore $\displaystyle 1=m^*(\bigcup_{i=1}^{\infty}V_i)$, however $$ \sum_{i=1}^{\infty}m^*(V_i)=\sum_{i=1}^{\infty} m^*(V)=\infty \times m^*(V). $$ This has two consequences. Firstly $m^*(V)$ cannot be zero because this will contradict the sub-additivity property of outer measures. Secondly because $m^*(V)\neq 0$ then $\sum_{i=1}^{\infty}m^*(V_i)=\infty$ which means: $$ \displaystyle 1=m^*(\bigcup_{i=1}^{\infty}V_i)<\sum_{i=1}^{\infty}m^*(V_i)= \infty $$

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  • $\begingroup$ Thanks a lot, this seems to clear the example. $\endgroup$ – Manjil P. Saikia Sep 7 '13 at 16:42

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