Give an example of a sequence $\{E_i\}$ of disjoint sets such that $m^*(\cup_{i=1}^{\infty}E_i)<\sum_{i=1}^{\infty}m^*(E_i)$.
Asked
Active
Viewed 1,206 times
0
-
Hi, I guess you mean pairwise-disjoint; just to be a bit more clear? – DBFdalwayse Sep 06 '13 at 07:26
-
Yes I meant pair wise disjoint. – Manjil P. Saikia Sep 06 '13 at 07:37
-
Have you seen the construction of a non-measurable set (say, from Royden's book)? – Prahlad Vaidyanathan Sep 06 '13 at 07:40
1 Answers
1
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: $$ \displaystyle 1=m^*(\bigcup_{i=1}^{\infty}V_i)<\sum_{i=1}^{\infty}m^*(V_i)= \infty $$
