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Let $a < c < d < b$. Let $f(x)=\begin{cases} 1 \text{ if } x \in [c,d]\\ 0 \text{ if } x \in [a,b]\setminus[c,d]. \end{cases}$

I want to verify that I understand how to properly apply the upper and lower Darboux sums.

Let $P=\{a=x_0<...<x_{k-1}<c<x_k<...<x_{l-1}<d<x_l<...<x_n=b\}$

$U(f,P)=\sum\limits_{i=1}^{n}M_i(f)\Delta x_i=\sum\limits_{i=k}^{l}M_i\Delta x_i=x_l-x_k \geq d-c.$

Verify that $L(f,P) \leq d-c$. This is my attempt at a verification.

$L(f,P)=\sum\limits_{i=1}^{n}m_i(f)\Delta x_i=\sum\limits_{i=k-1}^{l-1}m_i\Delta x_i=x_{l-1}-x_{k-1} \leq d-c.$

Is my verification correct?

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Shouldn't your $f(x)=0$ if $x\in [a,b]\setminus [c,d]$ instead of the other way around? – Stefan Hansen Jan 17 '13 at 12:08
Your definition of $f$ is not clear, as you use the interval $[c,d]$ both times. Also, since you use $\Delta x_i = x_i - x_{i-1}$ $$ L(f,P) = \sum_{i=k\color{red}{+1}}^{l-1}M_i\Delta x_i. $$ – Ilya Jan 17 '13 at 12:08
I just noticed that I defined the function incorrectly. Also, why does it have to be $k+1$ and not just $k$? – emka Jan 17 '13 at 12:11
@emka: well, it can be $k$ as well but the point is that $\min\{f(x):x\in [x_{k-1},x_k]\} = 0$ so that the sum ends up in $x_{l-1} - x_{k+1}$. – Ilya Jan 17 '13 at 12:13
@Ilya This makes more sense. Thanks! – emka Jan 17 '13 at 12:17
up vote 1 down vote accepted

Yes, this is correct except for some indices. Just to sum up: $$ M_i=\sup_{[x_{i-1},x_i]}f= \begin{cases}1,\quad &i=k,\ldots,l\\ 0,&\text{otherwise} \end{cases} $$ by the definition of $f$ and your partition $P$. A similar thing holds for $m_i$. Then $$ U(f,P)=\sum_{i=1}^n M_i(x_i-x_{i-1})=\sum_{i=k}^l (x_i-x_{i-1})=x_l-x_{k-1}\geq d-c, $$ and similarly for $L(f,P)$. Note that you want $x_l$ to be larger than $d$ but $x_{k-1}$ to be smaller than $c$ in order for the inequality to hold.

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Thanks for helping me clear this up. I haphazardly copied this example off of the board and didn't really understand it until now. – emka Jan 17 '13 at 12:17

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