Let *m* the n-dimensional Lebesgue measure on $R^n$.By definition of product measure, on each borelian set E


$m(E)=\inf \left(\sum\_{j=1}^\infty m(R\_j),\:\: E\subseteq \bigcup R\_j , \:\:R\_j \text{ rectangles}\right)$ 


It is also true that lebesgue measures are regular, so
$m(E)=\inf \left(m(U), E\subseteq U, \: U \text{ open set} \right)$.


Can I say that also holds
$m(E)=\inf \left(\sum\_{j=1}^\infty m(B\_j),\:\: E\subseteq \bigcup B\_j , \:\:B\_j \text{ balls}\right)$ or not? 



Lebesgue measure of a set
It follows from Vitali's covering theorem but not in an entirely trivial
fashion. We can reduce to the case where $E$ is open of finite measure.
The set of all open balls contained in $E$ is then a Vitali cover. By Vitali's
covering theorem there is a sequence of disjoint balls $(B\_n)$
whose union is a subset $U$ of $E$ with the same measure as $E$.
Thus $F=E-U$ is a set of Lebesgue measure zero.


Let $\epsilon>0$. There is a sequence of open cubes covering $F$
with total measure $<\epsilon$. Circumscribe these with balls
and we get a sequence of open balls covering $F$ with total measure $< c\_n\epsilon$
where $c\_n>0$. Interweaving this sequence with the $(B\_n)$ we get
a sequence of balls covering $E$ of total measure $< m(E)+c\_n\epsilon$.


