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I am trying to understand the proof of every monotonic function that is on an interval is integrable.

This is what I have $U(f, P) - L(f, P) = \sum\limits_{k=1}^n(f(t_k) - f(t_{k-1}))\cdot (t_k - t_{k-1})$

Now my book says that this is equal to:

$= (f(b) - f(a))\cdot(t_k - t_{k-1})$

How does one deduce that $\sum\limits_{k=1}^n(f(t_k) - f(t_{k-1})) = f(b) - f(a)$?

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It should be added that the interval is closed. –  Karolis Juodelė Dec 5 '12 at 7:37

2 Answers 2

up vote 1 down vote accepted

Note that \begin{equation*} \sum_{k=1}^{n}(f(t_{k})-f(t_{k-1}))=(f(t_{1})-f(t_{0}))+(f(t_{2})-f(t_{1}))+(f(t_{3})-f(t_{2}))+...+(f(t_{n})-f(t_{n-1})), \end{equation*} so in $i$:th term of the sum, $-f(t_{i-1})$ always eliminates the $i-1$:th appearing term $+f(t_{i-1})$. Hence you are only left the endpoints, i.e. $f(b)-f(a)$.

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You deduce that last identity because the sum telescopes. If you write $$ \sum_{k=1}^n (f(t_k) - f(t_{k-1})) = \sum_{k=1}^n f(t_k) - \sum_{k=0}^{n-1} f(t_k) = f(t_n) - f(t_0), $$ if $t_n = b$ and $t_0 = a$, this is what you want. The function must be integrable when $f$ is monotonic because $$ \left| \sum_{k=1}^n (f(t_k) - f(t_{k-1}))(t_k - t_{k-1}) \right| \le \max \{ (t_k - t_{k-1}) \} \sum_{k=1}^n |f(t_k) - f(t_{k-1})|. $$ By assumption on f monotonic, the absolute values get removed and all you keep is the sign of $f$ (i.e. + if $f$ increases, $-$ if it decreases). Therefore that sum telescopes to $f(b) - f(a)$, i.e. a constant that does not depend on the partition. Since the step goes to $0$ (the maximum in front of the sum), $f$ is integrable.

Hope that helps,

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