I N-dimensional Lebesgue measure: def. with Borel sets

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Let us define, as Kolmogorov-Fomin's 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...
 
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The opposite inequality ##\mu^*(A)\le\mu^*_B(A)## is proved via pretty standard reasoning.

Take arbitrary ##\varepsilon>0##. By the definition of ##\mu^*_B## there exists a cover of ##A## by Borel sets ##B_k##, ##k\ge1## such that $$\sum_{k\ge 1} m (B_k)<\mu^*_B(A)+\varepsilon.$$ The Borel sets are Lebesgue measurable, so for any such set ##m(B)=\mu^*(B)##. Therefore by the definition of ##\mu^*## every ##B_k## can be covered by parallelepipeds ##P_{k,j}##, ##j\ge 1##, such that $$\sum_{j\ge 1} m(P_{k,j} ) < \mu^*(B_k) + 2^{-k}\varepsilon= m(B_k) + 2^{-k}\varepsilon .$$ The parallelepipeds ##P_{k,j} ## cover ##A##, and $$\sum{k,j\ge 1} m(P_{k,j}) < \sum_{k\ge 1} m(B_k) + \sum_{k\ge 1}2^{-k}\varepsilon < \mu^*_B(A) +\varepsilon+ \varepsilon, $$ so $$\mu^*(A) \le \mu^*_B(A) + 2\varepsilon.$$ Since the last inequality holds for all ##\varepsilon>0##, we conclude that $$\mu^*(A) \le \mu^*_B(A) . $$ As for the measurability, it is even simpler. Borel sets are Lebesgue measurable, so if ##A## is approximated by a Borel set ##B##, ##\mu^*(A\Delta B)<\varepsilon##, then we can approximate ##B## by a simple set ##P##, ##\mu^*(B\Delta P)<\varepsilon##, so $$\mu^*(A\Delta P)<2\varepsilon.$$
 
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Very, very clear! I heartily thank you!!!
 

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