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Question about integral and unit step function
The unit step function $I$ is defined by
$ I(x)= \begin{cases}0,\quad x \le 0, \\ 1,\quad x>0. \end{cases} $
Let $f$ be continuous on $[a,b]$ and suppose $c_n\geq 0$ for $n=1, 2, 3,\ldots$ and $\sum_n c_n$ is convergent. Let $\alpha=\sum_{n=1}^{N} c_n I(x-s_n)$ where ${s_n}$ is a sequence of distinct points in $(a,b)$. Then $ \int_{a}^{b}fd\alpha=\sum_{i=1}^{N}c_n f(s_n). $
I asked this question yesterday and when I read other peoples answer it seems like that I understand well. But today I tried to prove this alone, it messed up again!
I drew $\alpha$ below.
As you can see, if n=1,2,...,N then $\alpha$ only has its value between ($s_n$,b] and it is $\sum c_n$.
I'm stuck in this point. Then $\int_{a}^{b} f d\alpha$ also has its value between ($s_n$,b]? If I put the $\alpha$ here, then should I integrate f with respect to constant value $\sum c_n$? Is it possible?
Actually, using the fact f is continuous on [a,b] is also considered. Then if x -> $s_n$ then f(x)->f($s_n$) but which part should I put this?
Please explain in detail so this time I can completely understand...