# Points of a Measure Zero Sets Covered by Intervals Infinitely Many Tmise

Given a measure zero set $E$, by definition we have forall $\varepsilon > 0$, a covering of $E$ by intervals whose lengths sum to $< \varepsilon$.

I want to prove that we can cover $E$ in intervals such that the sum of the lengths the intervals is $< 1$ and each point in $E$ is contained in infinitely many of the intervals.

Do you know how to prove this? Thank you

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Cover $E$ in intervals whose lengths sum to at most $\varepsilon$. Cover it again, but now with sum at most $\varepsilon/2$. Cover it again, but now with sum at most $\varepsilon/4$ … I think you begin to get the idea? Keep on doing this and take the union of all the covers, whose lengths sum to at most $\varepsilon$.

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Wonderful!!!!!! –  sperners lemma Oct 31 '12 at 20:25
But what if some of the intervals you choose at the $n$th step are the same as the intervals you chose at previous steps? –  Old John Oct 31 '12 at 20:27
@OldJohn, I think that's fine for the proof that uses this result. –  sperners lemma Oct 31 '12 at 20:28
@OldJohn: At any time in the construction, only a countable number of intervals have been selected. So you can always expand your chosen intervals by an arbitrarily small amount to avoid repetitions. –  Harald Hanche-Olsen Oct 31 '12 at 20:30
Yes, indeed - but I don't think that is necessary with my argument (I might be wrong!) –  Old John Oct 31 '12 at 20:31
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First find a cover of $E$ which has sum of lengths less than 1/2,
The find another cover of $E$ which has sum of lengths less than 1/4, etc. etc.
The union of all the covers will then be a cover with sum of lengths less than 1, and we just need to show that each point of $E$ is in an infinite number of the covering intervals (it might happen that some of the intervals chosen at each step are the same as intervals chosen at a previous step).
To prove this last point, take a point $x$ and suppose it is contained in only finitely many (say $n$) of the intervals. Let the length of the smallest of these be $y$. Then there is an interval in a "later covering" (of total length less than $y$) containing $x$, and it must be of length smaller than any of the $n$ intervals. This gives a contradiction.