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3
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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? |
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3
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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 $< m(E)+c_n\epsilon$. |
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