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 September 11th, 2014, 11:07 AM #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 measure-zero 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 measure-zero 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! Tags measure, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DuncanThaw Real Analysis 4 November 23rd, 2013 11:02 AM limes5 Real Analysis 4 July 26th, 2013 12:47 PM Jeh Real Analysis 0 July 19th, 2012 09:13 PM everk Real Analysis 1 October 18th, 2010 01:39 PM eskimo343 Real Analysis 1 December 6th, 2009 07:44 PM

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