TITLE: Controlling the Size of an Open Cover of a Set of Measure Zero
QUESTION [13 upvotes]: Suppose we have a subset $A\subset\mathbb{R}$ of Lebesgue measure zero contained in a compact interval, say $[0,1]$.  We know that since $A$ has measure zero we can cover $A$ with a countable set of open intervals, say $\{U_i\}$, such that $\mu(\cup_iU_i)\leq \varepsilon$ for any $\varepsilon$.  Now, if we fix some $\varepsilon>0$, can we cover $A$ with a countable set of open intervals, say now $\{V_i\}$, such that $\mu(V_i)\leq\frac{\varepsilon}{2^i}$ for each $i$?  That is, can we control the size of each individual set in some way? I have been thinking about this for a bit, and keep running into having to do things an infinite number of times, or having to choose the wrong indices first.  Any ideas?  
Thanks!

REPLY [12 votes]: If the property holds, then $A$ has Hausdorff dimension $0$, because $$\sum_{n=1}^\infty \left(\frac{\varepsilon}{2^n}\right)^d=\frac{\varepsilon^d}{2^d-1}$$ can be made arbitrarily small for each fixed $d>0$ by choosing $\varepsilon$ small enough.  The Cantor set has Lebesgue measure $0$ and Hausdorff dimension $\log_3(2)$, so it is a counterexample.