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I was working through Schilling's book on Measure theory 2ed and came across a problem and solution that I did not understand. Schilling published a solution guide, but I was not sure how he arrived at the solution, and I was hoping someone could fill in the gaps. It seems like the author is using some common tricks but he did not really explain the tricks. Any help is appreciated.

The question is 4.6 on page 29. The problem is:

find a measure $\mu$ on $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$ which is $\sigma$-finite but assigns to every interval $[a,b)$ with $b-a >2$ finite mass.

END QUESTION

The posted solutions as provided in the solution manual (page 35-36) is:

Define a measure $\mu$ which assigns every point $n - \frac{1}{2k}, \ n \in \mathbb{Z}, k \in \mathbb{N}$ the mass $\frac{1}{2k}$:

$$ \mu = \sum_{n \in \mathbb{Z}}\sum_{k\in \mathbb{N}}\frac{1}{2k}\delta_{n-\frac{1}{2k}} $$

Since $\mathbb{Z} \times \mathbb{N} $ is countably infinite so this is expression is indeed a valid measure. Any interval $[a,b)$ of length $b-a>2$ must contain an integer, say $m\in [a,b)$ such that $[m-1/2, m) \subset [a,b),$ thus by monotonicity:

$$ \mu[a,b) \geq \mu[m-1/2, m] = \sum_{k\in \mathbb{N}}\frac{1}{2k} = \infty $$

On the other hand, the sequence of sets $$ B_n := \bigcup_{k=-n}^n[k-1, k-\frac{1}{2n}] $$ satisfies $\mu(B_n) < \infty$ and $\cup_{n}B_n = \mathbb{R}$.

END SOLUTION

Note, the $\delta$ is a dirac delta function. The $\mathscr{B}(\mathbb{R})$ represents Borel sets over the Reals.

My questions:

  1. How did the author arrive at the expression for $\mu = \sum_{n \in \mathbb{Z}}\sum_{k\in \mathbb{N}}\frac{1}{2k}\delta_{n-\frac{1}{2k}}$?

I understand that this interval is supposed to represent a borel set. This seems like it a common trick of some sort. It actually reminds me of what Tao calls the $\frac{\epsilon}{2^n}$ trick that he employed in his measure theory text.

  1. Why does the solution construct the measure $\mu$ to sum over all integers and natural numbers?

This is a more specific extension of question 1 above. But why did the author set the measure to sum over all natural numbers and integers. This is basically going to lead to the measure going to infinity? Isn't there a simpler way to construct this measure that still meets the requirement of continuity from above and below.

  1. Why did he take the measure over the product of the integers and natural numbers?

This seems like a way to obtain a countably infinite number of sets--consistent with the axioms of a $\sigma$-algebra, but still cover the negative reals. Is my explanation sufficient or did I miss anything about this decision?

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    $\begingroup$ I guess it's a typo, and it should read "infinite mass"... $\endgroup$ – saz Aug 8 '17 at 13:48
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    $\begingroup$ There is a much simpler solution. Given $(\mathbb{R}, \mathscr{B}(\mathbb{R}))$, let us define $\mu$ by, for any $E \in \mathscr{B}(\mathbb{R})$, $$\mu(E)= \#(E\cap \mathbb{Q}) $$ (that is, $\mu(E)$ is the cardinality of $E\cap \mathbb{Q}$) Then $\mu$ is a $\sigma$-finite measure, and for any interval $[a,b)$ such that $a<b$, we have $\mu([a,b))=\infty$. In particular, for any interval $[a,b)$ such that $b-a>2$, we have $\mu([a,b))=\infty$. $\endgroup$ – Ramiro Aug 8 '17 at 14:00
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    $\begingroup$ @saz The book actually says "finite" mass. I also checked the list of errata for the book and it did not list this page. But I understand your point. $\endgroup$ – krishnab Aug 8 '17 at 17:49
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    $\begingroup$ @krishnab I know that the book says "finite", and I'm well aware of the list. However, the (second edition of the) book appeared rather recently, and therefore it is unlikely that the list contains all typos. Two reasons why I believe that it should read "infinite": 1. The example in the solution does assign infinite mass to the intervals. 2. For "finite mass" the problem would be trivial since we could take any finite measure $\mu$. $\endgroup$ – saz Aug 8 '17 at 17:58
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    $\begingroup$ @Ramiro Thanks for the alternative solution. This is definitely a simpler and more intuitive answer. I appreciate it. $\endgroup$ – krishnab Aug 9 '17 at 2:22

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