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N-dimensional Lebesgue measure: def. with Borel sets
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[QUOTE="Hawkeye18, post: 5502187, member: 259386"] 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 $$\mu^*(A\Delta P)<2\varepsilon.$$ [/QUOTE]
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N-dimensional Lebesgue measure: def. with Borel sets
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