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A well-known result is that we can always construct a countably additive function $\mu$ from a nondecreasing and right-continuous function $G$. More specifically, we define on the semiring $\mathcal{C}$ of all intervals $(a,b]$, $$\mu((a,b])=G(b)-G(a),$$ where $\mu$ is the Lebesgue measure when $G$ is the identity mapping. I'm curious if the following function gets this property as well:

Fix a countable set $C=\{c_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where each $c_n$ is distinct, and a countable set $\{a_n:n\in\mathbb{N}\}\subset\mathbb{R}$ where each $a_n$ is non-negative and $\sum_na_n<\infty$.

Define $G:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x\longmapsto\sum\{a_n:c_n\leq x\}.$$

That is, is this function right-continuous and non-decreasing? Furthermore, does the measure $\mu$ constructed from this function satisfy $\mu(\{c_n\})=a_n$ and $\mu(\mathbb{R}\backslash C)=0?$

EDIT: $G$ being right-continuous and non-decreasing seems pretty easy to see. Non-decreasing is rather obvious from the property of the $a_n$'s and I basically argued right-continuity to myself in the comment box below.

However for the second part, I'm not sure how to approach either property. How do you interpret $\mu(\{c_n\})$ to get only $a_n$ remaining in the subtraction? I appreciate any help!

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I'm not sure I fully understood, but unless there is an order-preserving map between $a_n$ and $c_n$ , I don't see how the map can be right-continuous; it seems like it would jump all-over the place. – gary Nov 15 '11 at 5:08
Note that the function/summation remains constant for all $x$ where $c_n\leq x < c_{n+1}$. It then "jumps" to the next interval when $x=c_{n+1}$. That is, the function is basically just intervals jumping to the next, where each interval is closed on the left. (as in Actually, I'm not too sure if this argument works because I'm assuming that the $c_n$'s are non-decreasing. Could I somehow build an analogous argument by re-ordering the $c_n$'s or doing something similar? – Dustin Tran Nov 15 '11 at 5:25
Actually, it does seem this argument is still valid. I can just replace $c_{n+1}$ with the condition that it holds for every $c_i>c_n$. That is, the function remains constant for all $x$ where $c_n\leq x<c_i$ for all $c_i>c_n$. Then the constant interval would jump to another when $x$ reaches one of these $c_i$'s. – Dustin Tran Nov 15 '11 at 5:36
Have you seen… ? – Byron Schmuland Nov 15 '11 at 15:22
About arguments based on reordering the $c_n$'s, keep in mind that $\{c_n\mid n\in\mathbb N\}$ might be a set like $\mathbb Q$ the set of rational numbers. How does one reorder $\mathbb Q$? – Did Nov 18 '11 at 6:43
up vote 1 down vote accepted

Some hints: Using indicator functions write $$G(x)=\sum_n a_n1_{[c_n,\infty)}(x).$$ We don't assume that the $c_n$s are ordered, indeed they may well be dense in $\mathbb{R}$. Nevertheless $G$ is right continuous since it is the uniform limit of $G_N(x)=\sum_{n=1}^N a_n1_{[c_n,\infty)}(x)$, which are obviously right continuous. The uniform convergence uses the summability of the sequence $a_n$.

For $\varepsilon>0$, we have $G(c_n)-G(c_n-\varepsilon)=\sum_m a_m$, where the sum is over all $m$ with $c_n-\varepsilon<c_m\leq c_n$. Even though the set of such $c_m$ may be infinite for every $\varepsilon>0$, the value of $\sum_m a_m$ decreases to $a_n$ as $\varepsilon\downarrow 0$.

Thus, we have $\mu(\{c_n\})=G(c_n)-G(c_n-)=a_n$ for each $n$. Since $\mu(C)=\mu(\mathbb{R})=\sum_n a_n$, we conclude that $\mu(\mathbb{R}\backslash C)=0$.

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Other than using my "interval jumping argument", what's a clearer way to explain that $G_N(x)=\sum_{n=1}^N a_n1_{[c_n,\infty)}(x)$ is right-continuous? It seems that when I try to explain that this is right-continuous, it uses the same logic as showing that the desired function is right-continuous in the first place. I am a fan of that clever use of the indicator function though! – Dustin Tran Nov 15 '11 at 9:07
What I'd do is convince myself that, by definition, $a 1_{[c,\infty)}$ is right continuous. Then show that the finite sum of right continuous functions is again right continuous. – Byron Schmuland Nov 15 '11 at 13:25

Let $a=\sum\limits_na_n$. Since $\{c_n\}$ is the decreasing intersection of the sets $]c_n-1/k,c_n]$ and the measure of these is finite, $\mu(\{c_n\})$ is the decreasing limit of $$ G(c_n)-G(c_n-1/k)=\sum\limits_ia_i\cdot[c_n-1/k\lt c_i\leqslant c_n]. $$ When $k\to\infty$, $a_i\cdot[c_n-1/k\lt c_i\leqslant c_n]\to a_n\cdot[i=n]$ hence Lebesgue convergence theorem shows the RHS converges to $a_n$.

Likewise, $\mu(\mathbb R)=a$ by definition and, by countable additivity, $$ \mu(C)=\sum\limits_n\mu(\{c_n\})=\sum\limits_na_n=a, $$ hence $\mu(\mathbb R\setminus C)=0$.

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Thanks for the clear explanation! That made perfect sense. – Dustin Tran Nov 15 '11 at 9:14

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