Rudin PMA 4.31 - does the elements of $E$ have to be ordered (smallest to the biggest)?

Question

Here is a reformulation of Rudin PMA $4.31$ remark:

Let $a$ and $b$ be two real numbers such that $a < b$, let $E$ be any countable subset of the open interval $(a,b)$, and let the elements of $E$ be arranged in a sequence $$x_1, x_2, x_3, \ldots.$$ Now let $\{c_n\}$ be any sequence of positive real numbers such that the series $\sum c_n$ converges. Now define the function $f \colon (a,b) \to \mathbb{R}$ as follows: $$f(x) \colon= \sum_{x_n < x} c_n \ \ \ \ \text{ for all } x \in (a,b).$$ The summation is to be understood as follows: Sum over those indices $n$ for which $x_n < x$.

I am wondering whether the $x_1, x_2,x_3,...$ from $E$ are necessarily ordered or not ?

I ask because, next in the remark, Rudin asserts that "$f(x_n +)-f(x_n -) = c_n$". However (and even though I haven't yet found a proof of this result), when drawing a sketch I've noticed that this assertion holds, only if $x_n$ is monotonically increasing (that is to say, if the elements of $E$ are arranged in $x_n$ in an ordered way).

When they are ordered in a reversed order ($(x_1,x_2,x_3,...) = (0,-1,-2,...)$ for example) I found (again with my graph) that $$f(x_n -) - f(x_n +)= c_{n+1}$$

hold instead.

Am I completely wrong, or Rudin is indeed considering only the monotonically increasing $x_n$ sequences ?

Answer 1

Consider this example: $(a,b) = (0,1)$ with $x_1= 3/4$, $x_2=1/2$, $c_1 = 1$, and $c_2=2$. Then \begin{equation*} f(x) = \begin{cases} 0, & x \leq \frac{1}{2}\\ 2, & \frac{1}{2} < x \leq \frac{3}{4}\\ 3, & \frac{3}{4} <x \end{cases} \end{equation*}

It seems to me that the order is irrelevant as noted by Rudin.

Note The comment below of krm2233 gives a nice explanation of how this example is constructed from the definition

Answer 2

Let $[x_n < x]$ be the indicator function on $x_n<x$. Then $$ f(x) = \sum_{n=1}^\infty c_n[x_n < x] $$ Now by checking cases we see that $$ f(x_k+\epsilon) - f(x_k-\epsilon) = \sum_{n=1}^\infty c_n[x_k-\epsilon\leq x_n<x_k+\epsilon] $$ as $\epsilon \searrow 0$ each $n\neq k$ is ommited from the above sum. Hence, in the limit the only number that remains is $c_k$. So the order is really irrelevant.

Answer 3

It may help to consider the auxiliary functions $f_m(x)$, where $$f_m(x) = \sum_{x_n<x, n\leq m} c_n$$.

Since the series $\sum c_n$ has positive terms, for fixed $x$, the sequence $f_m(x)$ is weakly increasing and converges (since is it bounded above by $\sum_n c_n$) with the limit being $f(x)$.

Now if $x = x_N$ say, then for each $m$ there is a $\delta_m>0$ such that $$0<\delta_m< \min\{|x_N-x_k|: 1\leq l\leq m, k \neq N\}.$$

If $m<N$ then $f_m(x)$ is constant on $(x_N-\delta_m,x_N+\delta_m)$ while if $m \geq N$ then $f_m$ is constant on $(x_N-\delta_m,x_N)$ and on $[x_N,x_N+\delta_m)$, where if $c=f(x)$ for $x \in (x_N-\delta_m,x_N)$ then $f(x) = c+c_N$ for $x\in [x_N,x_N+\delta_m)$, and thus if $m\geq N$ we have $f_m(x_N+)-f_m(x_N-) = c_N$.

It is not too hard to deduce from this that $f(x_N+)-f(x_N-) = c_N$...

Answer 4

I do not think that it is assumed that the $x_i$ are ordered, and that is not needed to assert that $$ f(x_n+)-f(x_n-)=c_n. $$ Given $n$, $f(x_n+)=\sum_{k:x_k\le x_n} c_k$ because any contribution from $x_k\in (x_n,x_n+\varepsilon)$ will eventually (as $\varepsilon\to 0$) vanish. On the other hand, $f(x_n-)=\sum_{k:x_k< x_n} c_k$ because all the contributions from $x_k\in (x_n-\varepsilon,x_n)$ will be eventually included. Therefore, the jump at $x_n$ is $$ \sum_{k:x_k\le x_n} c_k-\sum_{k:x_k< x_n} c_k=c_n $$