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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 can't understand why the last equation holds. Where does $f(s_n)$ comes from?

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Hint: consider first $N=1$. – Siminore Nov 20 '12 at 12:29
up vote 2 down vote accepted

Remember the definition of the Riemann-Stieltjes integral. In the present example, the integrator $\alpha$ is constant between two consecutive $s_n$, hence two consecutive points from a partition of $[a,b]$ only contribute to the corresponding Riemann-Stieltjes sum if there is a $s_n$ between them, i. e. if $\alpha$ has a jump in between them. Their contribution is exactly the sum of all jumps happening. If the partition is fine enough, at most one jump happens between two neighboring points of the partition.

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I drew this function and figure out what you mean! Thanks! – niagara Nov 20 '12 at 12:33

$$ \begin{eqnarray} \int_{a}^{b}f(x)d\alpha(x)&=&\int_{a}^{b}f(x)d\left(\sum_{n=1}^{N} c_n I(x-s_n)\right)\\ &=&\sum_{n=1}^{N}c_n\int_{a}^{b}f(x)d\left( I(x-s_n)\right)\\ &=&\sum_{n=1}^{N}c_n\int_{a}^{b}f(x)\frac{d\left( I(x-s_n)\right)}{dx}dx\\ &=&\sum_{n=1}^{N}c_n\int_{a}^{b}f(x)\delta(x-s_n)dx\\ &=&\sum_{i=1}^{N}c_n f(s_n) \end{eqnarray} $$

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