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Construction of a Borel set with positive but not full measure in each interval

I came across the following question in my self-study:

Where $\mu$ is the Lebesgue measure, show there exists a Borel set $A \subseteq [0,1]$ such that $0 < \mu(A \cap I) < \mu(I)$ for every subinterval $I \subseteq [0,1]$. (Hint from the book I am consulting: Every subinterval of $[0,1]$ contains Cantor-type sets of positive measure.)

This result (and its hint) seem quite interesting, and I have attempted a few times to lasso a proof of this result without success, and am curious to see if anyone visiting is up for proving this interesting result.

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    Think about constructing a positive measure Cantor set. Think along the analogy of the middle-third Cantor set, where you remove at each step the middle third of each interval. Now, instead of removing mid-third at each step, remove successively smaller middle parts. For details, see: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set2012-01-28

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