September 11th, 2014, 12:07 PM  #1 
Member Joined: Mar 2013 Posts: 71 Thanks: 4  problem about measure
Please, check my solution to this problem: "Prove that if $X \subset [a,b]$ isn't a measurezero set, then there exists $\varepsilon >0$ such that, for every partition $P$ of $[a,b]$, the sum of the lenghts of the intervals of $P$ which contain points of $X$ is greater than $\varepsilon$". Solution: Let $X \subset [a,b]$ be a non measurezero set. Then there exists $\varepsilon>0$ and an enumerable family of open intervals $I_1,...,I_n,...$ such that $X \subset I_1 \cup ... \cup I_n \cup ...$ with $\sum_{i=1}^{\infty} I_i\ge \varepsilon$. Without lost of generality, We can supose $I_1 \cup ... \cup I_n \cup ... \subset [a,b]$. In such a case, the points $a$, $b$ and the extremities of the intervals $I_1,...,I_n,...$ form a partition of $[a,b]$. By construction, the intervals of this partition which contain points of $x$ have length $ \ge \varepsilon$. Thank you for your attention! 

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