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Let's say I have two fixed countable sets, $A=\{a_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where the $a_n$ are all distinct, and $\{b_n:n\in\mathbb{N}\}\subset\mathbb{R}$. Here $b_n\geq 0$ and $\sum_n b_n<\infty$. I define $F\colon\mathbb{R}\to\mathbb{R}$ by $$ F(x)=\sum \{b_n: a_n\leq x\}. $$

It's not hard to see that this function is nondecreasing, and can be shown the function is continuous on the right.

It's a standard theorem of measure theory that the function $\mu((a,b])=F(b)-F(a)$ on the collection of half-open intervals $(a,b]$ for $a\leq b$ can be extended to a unique measure (also denoted $\mu$) on the Borel $\sigma$-algebra of $\mathbb{R}$.

But what's the reason why this measure takes values on singletons by $\mu(\{a_n\})=b_n$, and if I remove the set $A$, then I get a set with measure $0$, i.e., $\mu(\mathbb{R}\setminus A)=0$? Thanks.

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You can take infinte intersections of ever smaller half-open intervals, so that in the limit you only have one point. –  Raskolnikov Nov 11 '11 at 21:56

1 Answer 1

up vote 1 down vote accepted

If you can prove the result on the singletons, the measure of $\mathbb{R}\setminus A$ will follow, since: $$\mu(\mathbb{R})=\sum_{n\in\mathbb{Z}}\mu(n,n+1] = \sum_{n\in\mathbb{N}}b_n,$$ and by $\sigma$-additivity we have: $$\mu(\mathbb{R}\setminus A) = \mu(\mathbb{R}) - \sum_{n\in\mathbb{N}}\mu(\{a_n\})= \sum_{n\in\mathbb{N}}b_n - \sum_{n\in\mathbb{N}}b_n = 0.$$

To see that $\mu\bigl(\{a_n\}\bigr) = b_n$, note that if $a_n\in (a,b]$, then $$\begin{align*} \mu((a,b]) &= F(b)-F(a)\\ &= \sum\{b_m:a_m\leq b\} - \sum\{b_m: a_m\leq a\}\\ &= \sum\{b_m: a\lt a_m\leq b\}\\ & \geq b_n.\end{align*}$$ Therefore, since $\mu(\{a_n\})$ is the infimum of the measure of any set $(a,b]$ that contains it, it follows that $\mu(\{a\}) \geq b_n$.

On the other hand, let $\epsilon\gt 0$.

Then there exists $N\gt 0$ such that $\sum_{m\geq N}b_m\lt \epsilon$. Let $\delta\gt0$ be such that $(a_n-\delta,a_n]$ does not contain any of $a_1,\ldots,a_{N-1}$ (except possibly $a_n$ itself). This is possible, since we only have to avoid finitely many points of $\mathbb{R}$. Then: $$\mu(\{a_n\}) \leq \mu\bigl((a_n-\delta,a_n]\bigr) = \sum_{a_n-\delta\lt a_m\leq a_n+\delta}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!b_m \ \ \ \ \ \leq b_n+\sum_{m\geq N}b_m = b_n+\epsilon.$$

Thus, $b_n\leq \mu(\{a_n\}) \leq b_n+\epsilon$ for every $\epsilon\gt 0$, which means that $\mu(\{a_n\}) = b_n$, as desired.

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Thanks! One question, after the align environment, is $\mu(\{a\})$ supposed to be $\mu(\{a_n\})$? Otherwise I don't see how $b_n\leq\mu(\{a_n\})$ in the last line. –  Vika Nov 15 '11 at 9:59
    
@Vika: Yes, it's meant to be $a_n$. –  Arturo Magidin Nov 15 '11 at 14:48

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