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I am trying to figure out a practice problem: Let $I_j$ be intervals in $[0,1]$ whose measure is $> \epsilon > 0$. Show that there is a subsequence $I_{jk}$ such that the measure $L(\cap I_{jk}) >0$.

So far I have shown that the intersection must be non-empty. If it is equal to zero, I assume that it must be a collection of points, but would like any tips/guidance to get me back on track.

smoothFunctional00
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  • They all have the same positive measure? So not $L(I_n) = 2^{-n}$ for example? – GEdgar Jun 03 '21 at 20:53
  • This is false as written -- consider any convergent series whose sum is less than one and convert those to consecutive intervals in $[0,1]$ (so the boundaries of these intervals are the partial sums of the series). It has some hope of being true if "whose measure is ${}>0$" is replaced with "whose measures are ${}> \varepsilon$ for some $\varepsilon > 0$." – Eric Towers Jun 03 '21 at 20:57
  • If instead you assume that there is a fixed number $\varepsilon\gt0$ such that all the intervals have length $\gt\varepsilon$, then it's true and quite easy. Cover $[0,1]$ with a finite collection $J_1,\dots,J_N$ of intervals of length $\lt\frac{\varepsilon}3$. Each interval in your sequence must contain one of the intervals $J_i$, and infinitely many of them contain the same one. – bof Jun 03 '21 at 21:08
  • For a counterexample where the $I_n$ are not pairwise disjoint, you can take $I_n = (0, 1/n)$. –  Jun 03 '21 at 21:16
  • @EricTowers good find, i edited the question – smoothFunctional00 Jun 04 '21 at 03:35
  • Of possible relevance (deals with arbitrary measurable sets, not intervals), see the comments to Pigeonhole principle in terms of measure of sets, especially sci.math post cited (which, as I have time, is being slowly being rewritten, TeX'ed, and expanded for an answer there). – Dave L. Renfro Jun 04 '21 at 05:03

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