3
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ Yes, this is Vitali's covering theorem. $\endgroup$ Mar 27, 2010 at 17:55
  • $\begingroup$ Typo? Did you really mean for the upper bound of each summation to be n? $\endgroup$ Mar 27, 2010 at 18:47
  • $\begingroup$ In fact, the upper bounds in sums should be removed. You need countable coverings in both cases, finite ones are not enough. $\endgroup$ Mar 27, 2010 at 20:58
  • $\begingroup$ Yes, typo, sorry $\endgroup$
    – Nicolò
    Mar 28, 2010 at 0:17

1 Answer 1

3
$\begingroup$

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 $< m(E)+c_n\epsilon$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.