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I am trying to understand how these lecture notes:https://www.mat.univie.ac.at/~armin/lect/real_analysis.pdf construct lebesgue measure on page 7 and I am stuck on part 2 of theorem 2.1. Could someone please clarify the whole proof of that part? To be more specific I would like to know why can we take the diameter of each interval $Q_i$ in $R$ to be less than $\delta$ after subdividing and also how did he combine the last two infinite series into one to get the required inequality he was looking for.

HAT
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  • It is not at all clear what part of the proof you do not understand, nor what form you hope an answer to be. What sort of clarification are you looking for? – SlipEternal Nov 25 '19 at 21:24
  • @InterstellarProbe basically why we can take the diameter of each of the closed intervals (I am thinking in R here) less than delta and also the inequality for the infinite series in the last part (when he combines the two infinite series into one on the right) are the two parts I am stuck which is essentially the whole proof. – HAT Nov 25 '19 at 21:26

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Suppose there exists an interval with a diameter greater than $\delta$. Subdivide it by cutting it in half. Example, suppose $\delta = 0.00001$ and you have an interval $[4,5]$. You can write that as $[4,4.5]\cup [4.5,5]$, then $[4,4.25],[4.25,4.5],[4.5,4.75],[4.75,5]$, etc. And eventually, you will get every interval to have a diameter less than $\delta$. Essentially, given any $\delta>0$ and any $x \in \mathbb{R}^+$, there exists $n \in \mathbb{N}$ such that $\dfrac{x}{2^n} < \delta$. This implies a countable number of subdivisions to make sure that every interval has diameter less than $\delta$ (the union of a countable collection of countable sets is countable). Here is a proof:

Prove that the union of countably many countable sets is countable

He combined them by the same principle. If you have a countable collection of sets, then there exists a bijection between the countable collection and an index set that uses only positive even numbers. Similarly, there exists a bijection between the countable collection and an index set that uses only positive odd numbers. So, if you want to add them up together, you can sum over the index set of all positive natural numbers (odds and evens together gives all natural numbers). Basically, it is just ways to manipulate countable collections into more easily manipulable forms.

SlipEternal
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  • such division was needed so that the new intervals resulting from such subdivision to have intersection with only one of the $E_i$ right? – HAT Nov 25 '19 at 21:46
  • Correct, because you are given that the minimum distance between points of $E_1$ and $E_2$ is at least $\delta>0$, a positive finite real number. – SlipEternal Nov 25 '19 at 21:47
  • Thanks and one last thing that last inequality is still bugging me, how do we ensure such an rearrangement and taking sums of terms as in this situation satisfies this inequality? – HAT Nov 25 '19 at 21:56
  • All of the $Q_j$'s are from the original cover of $E_1\cup E_2$. Taking subdivisions does not change the total measure (as shown in my example where I divided up an interval, but the outer measure of each interval adds to the same total as the original interval). Basically, in $\mathbb{R}^n$, you can cut each "cube" in half, and the intersection of the two will have outer measure zero. – SlipEternal Nov 25 '19 at 22:05
  • And then we use commutativity of addition to be able to rearrange them. – SlipEternal Nov 25 '19 at 22:07