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

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1. Jun 19, 2016

### DavideGenoa

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...

2. Jun 19, 2016

### Hawkeye18

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 $$\mu^*(A\Delta P)<2\varepsilon.$$

3. Jun 19, 2016

### DavideGenoa

Very, very clear! I heartily thank you!!!