# Darboux Sums

• May 14th 2011, 03:36 PM
Conn
Darboux Sums
So we were doing Darboux sums in Analysis, and this was a given sample question and answer, but I don't really understand where part of this solution comes from:

"Suppose f is increasing on [0,1]. Show that f is integrable on [0,1]

Let P = $\displaystyle \left\{ x_{0},x_{1},...,x_{n} \right\}$ be a partition of [0,1] into n subintervals of the same length.
Since f(x) is increasing, we have
$\displaystyle \inf_{[x_{k+1},x_{k}]} f(x) = f(x_{k}) for 0 \leqslant k \leqslant n-1$
and this implies that

$\displaystyle S^{-}(f,P) = \sum_{k = 0}^{n-1} \inf_{[x_{k+1},x_{k}]} f(x).(x_{k+1}-x_{k})$

$\displaystyle = \frac{1}{n}(f(x_{0})+f(x_{1})+...+f(x_{n-1})$

But I don't understand how those last two statements are equal; for example, how is

$\displaystyle \sum_{k = 0}^{2} f(x).(x_{k+1}-x_{k})$

$\displaystyle = (f(x_{0})(x_{1}-x_{0})+f(x_{1})(x_{2}-x_{1})+f(x_{2})(x_{3}-x_{2}))$

equal to

$\displaystyle \frac{1}{3} (f(x_{0})+f(x_{1})+f(x_{2})?$
• May 14th 2011, 03:48 PM
TheEmptySet
Quote:

Originally Posted by Conn
So we were doing Darboux sums in Analysis, and this was a given sample question and answer, but I don't really understand where part of this solution comes from:

"Suppose f is increasing on [0,1]. Show that f is integrable on [0,1]

Let P = $\displaystyle \left\{ x_{0},x_{1},...,x_{n} \right\}$ be a partition of [0,1] into n subintervals of the same length.
Since f(x) is increasing, we have
$\displaystyle \inf_{[x_{k+1},x_{k}]} f(x) = f(x_{k}) for 0 \leqslant k \leqslant n-1$
and this implies that

$\displaystyle S^{-}(f,P) = \sum_{k = 0}^{n-1} \inf_{[x_{k+1},x_{k}]} f(x).(x_{k+1}-x_{k})$

$\displaystyle = \frac{1}{n}(f(x_{0})+f(x_{1})+...+f(x_{n-1})$

But I don't understand how those last two statements are equal; for example, how is

$\displaystyle \sum_{k = 0}^{2} f(x).(x_{k+1}-x_{k})$

$\displaystyle = (f(x_{0})(x_{1}-x_{0})+f(x_{1})(x_{2}-x_{1})+f(x_{2})(x_{3}-x_{2}))$

equal to

$\displaystyle \frac{1}{3} (f(x_{0})+f(x_{1})+f(x_{2})?$

It follows from the fact that

$\displaystyle (x_{k+1}-x_k)=\frac{1}{n}, \quad \forall \, k$

Then just factor it out of each term!