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I am reading "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

What is $\alpha$?
I can imagine what $\beta(x)=\sum_{n=1}^{N} c_n I(x-s_n)$ is.
We can assume that $s_1<s_2<\dots<s_N$.
$\beta$ looks like the following picture: enter image description here

But when $N=\infty$, I cannot imagine what $\beta=\alpha$ is.

$\alpha$ can be continuous?
$\alpha$ can be differentiable?

6.14 Definition The unit step function $I$ is defined by $$I(x)=\begin{cases}0 & (x \leq 0),\\ 1 & (x > 0).\end{cases}$$

6.16 Theorem Suppose $c_n\geq 0$ for $n=1,2,3,\dots$, $\sum c_n$ converges, $\{s_n\}$ is a sequence of distinct points in $(a,b)$, and $$\alpha(x)=\sum_{n=1}^\infty c_n I(x-s_n).$$
Let $f$ be continuous on $[a,b]$. Then $$\int_{a}^{b}f\,d\alpha=\sum_{n=1}^\infty c_nf(s_n).$$

tchappy ha
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    It is not possible to draw the graph of $\alpha$. $(s_n)$ could be the set of all rational numbers, for example. – Kavi Rama Murthy Apr 04 '22 at 07:27
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    This theorem looks a bit strange, are you sure you got all the details right in your question? For example what stops me from setting $c_i = 0$ everywhere and then the integral is always zero? Which of course can not be true for all functions $f$. I notice that $\alpha$ is defined in the theorem but not used at all. – Cristian Gratie Apr 04 '22 at 07:30
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    @CristianGratie I am very poor at English, so I cannot understand what you are saying exactly. $\int_{a}^{b} f d\alpha$ is not a Riemann integral. It is a Riemann - Stieltjes integral. – tchappy ha Apr 04 '22 at 07:41
  • @KaviRamaMurthy Thank you very much for your comment. – tchappy ha Apr 04 '22 at 07:42
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    In more detail, we have to imagine a bounded, non-decreasing function with a jump discontinuity of size $c_n$ at $s_n$ for each $n$. But that at least tells us $\alpha$ is discontinuous (unless all the $c_n$ are zero), generally at infinitely many points, and possibly at each point of a dense set. – Andrew D. Hwang Apr 04 '22 at 11:13
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    @tchappyha Ah, I read $dx$ instead of $d\alpha$, my bad. If you want to visualize the example you started on paper, that is also possible. You just need to notice that $s_i < b$, so as you go to the right, your $s_i$ values will be closer and closer, so the distance between $s_i$ and $s_{i+1}$ (the horizontal segments in your drawing) will get shorter and shorter. Concrete example you can draw: $s_n = 1 - 2^{-n}$, $c_n = 2^{-n}$ for $n > 0$ and $c_0=0$. With this you will notice $\alpha(s_n) = s_n$ and it will look like a step approximation of $f(x) = x$ with smaller steps as you approach $1$. – Cristian Gratie Apr 04 '22 at 11:43
  • @AndrewD.Hwang Thank you very much for your comment. – tchappy ha Apr 08 '22 at 08:01
  • @CristianGratie Thank you very much for your comment. – tchappy ha Apr 08 '22 at 08:01

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