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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 en.wikipedia.org/wiki/File:Right-continuous.svg) 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 math.stackexchange.com/questions/81218/… ? –  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

2 Answers 2

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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