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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.

enter image description here

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...

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I don't think it is a good idea to re-ask the very same question you asked yesterday. Better, go back to the first question and ask again, under answers or wherever, about doubts and problems you still have. –  DonAntonio Nov 21 '12 at 14:50
    
okay,then how can I delete this question? –  niagara Nov 21 '12 at 14:51
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marked as duplicate by DonAntonio, The Chaz 2.0, Nate Eldredge, Martin Argerami, Did Nov 21 '12 at 15:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer

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As suggested in a comment to your related question, let's consider the case $N = 1$ first. Then $\alpha = c_1I(\cdot - s_1)$. Now let $\Delta = \{a = x_0 < \ldots < x_k = b\}$ a partition of $[a,b]$ and $\xi_i \in [x_i, x_{i+1})$ for $i < k$. The integral is the limit of sums \begin{align*} S(\Delta, (\xi_i)) &= \sum_{i=0}^{k-1} f(\xi_i)\bigl(\alpha(x_{i+1}) - \alpha(x_i)\bigr) \end{align*} Now: $\alpha$ can take only two values, namely $c_1$ and $0$ and there is exactly one $i$ where $\alpha(x_i) \ne \alpha(x_{i+1})$, namely when $x_i \le s < x_{i+1}$ Hence $$ S(\Delta, (\xi_i)) = c_1f(\xi_i), \qquad s,\xi_i \in [x_i, x_{i+1}) $$ Now, for $|\Delta| \to 0$, we have $|x_{i+1}- x_i| \to 0$, hence $|s - \xi_i| \to 0$. As $f$ is continuous, we have $f(\xi_i) \to f(s)$. This gives $$ S(\Delta, (\xi_i)) \to c_1f(s)= \int_a^b f\,d\alpha $$ Now note, that $\alpha \mapsto \int_a^b f\, d\alpha$ is linear to obtain your result.

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Thanks a lot! I almost forgot the concept of Riemann sum. Thank you for your comment! –  niagara Nov 21 '12 at 15:08
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