- #1

DavideGenoa

- 155

- 5

*Элементы теории функций и функционального анализа*does, the definition of

*outer measure*of a bounded set ##A\subset \mathbb{R}^n## as $$\mu^{\ast}(A):=\inf_{A\subset \bigcup_k P_k}\sum_k m(P_k)$$where the infimum is extended to all the possible covers of ##A## by finite or countable families of ##n##-parallelipeds ##P_k=\prod_{i=1}^n I_i##, where ##I_i\subset\mathbb{R}## are finite intervals (where one point may be considered an interval), whose measure ##m(P_k)## is the product of the lengths of the intervals #I_i#.

A set is said to be

*elementary*when it is the union of a finite number of such ##n##-parallelipeds and a set ##A## is said to be *measurable* if, for any ##\varepsilon>0##, there is an elementary set ##B## such that $$\mu^{\ast}(A\triangle B)<\varepsilon.$$The function ##\mu^{\ast}## defined on measurable sets only is called

*Lebesgue measure*and the notation ##\mu## is used for it.

I have been told that, in the definitions explained above, we can equivalently use Borel sets where ##n##-paralallelepipeds appear.

How can we prove it?

Let us use the index ##B## for such alternative definitions. It is clear to me that, for any set ##A\subset \mathbb{R}^n##, $$\mu_B^{\ast}(A)\le \mu^{\ast}(A)$$ because ##n##-parallelepipeds are Borel sets, and if ##A## is measurable according to the ##n##-parallelepipeds-based definition, it also is according to the Borel-sets-based definition, but I cannot prove the converse...