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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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    @OldJohn: No, your argument is perfectly fine. (I didn't even see your argument the previous time around, since following the link to the comment from my inbox scrolled your answer off the page. I need to become more aware of this.)2012-10-31
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How about this:

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.