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Hey, I tried reading over this in one of my classes today since I fell way behind in all my work. Any ideas?
Let f: [a,b] --> R be increasing on the set [a,b] (i.e., f(x) <= f(y) whenever x<y). Show that f is integrable on [a,b].
I missed these lectures and the textbook is so cryptic.
what does this have to with f having discontinuities? start by finding formulas for the uper and lower sums. the idea is you want to show that their difference in absolute values can be made less than any $\displaystyle \epsilon > 0$Quote:
Originally Posted by ajj86
For any partition $\displaystyle P=\left\{ {{x}_{0}},{{x}_{1}}\ldots ,{{x}_{n}} \right\}$ of $\displaystyle [a,b],$ it's $\displaystyle m_i=f\big(x_{i-1}\big)$ and $\displaystyle M_i=f\big(x_i\big).$ (Where $\displaystyle m_i$ and $\displaystyle M_i$ denote the infimum and supremum of $\displaystyle f$ in the interval $\displaystyle [x_{i-1},x_i].$)
Take a $\displaystyle P$ partition which has its $\displaystyle n$ equal subintervals, then $\displaystyle S(P)-s(P)=\sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)\frac{b-a}{n}}-\sum\limits_{i=1}^{n}{f\left( {{x}_{i-1}} \right)\frac{b-a}{n}}=\big(f(b)-f(a)\big)\frac{b-a}{n},$ where $\displaystyle S(P)$ and $\displaystyle s(P)$ are the upper and lower sum of $\displaystyle f$ in the given partition. Finally, given $\displaystyle \epsilon>0,$ just take $\displaystyle n>\frac{(b-a)\big(f(b)-f(a)\big)}\epsilon$ and we get $\displaystyle S(P)-s(P)<\epsilon,$ which means that $\displaystyle f$ is integrable on $\displaystyle [a,b].$
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There's another theorem which talks about discontinuities:
"Let $\displaystyle f:[a,b]\to\mathbb R$ be a bounded function which only has a finite number of discontinuities on $\displaystyle [a,b],$ then $\displaystyle f$ is integrable on $\displaystyle [a,b].$"
