# If $f$ is increasing and continuous on $[a,b]$, then it is integrable there.

I am trying to prove the following:

Show that an increasing, continuous function $f$ on $[a,b]$ is integrable there.

This is my idea:

Let $f$ be an increasing, continuous function on $[a,b]$. Take a partition $P$ of $[a,b]$ of $n$ equal-length sub-intervals. Then \begin{align} U(f,P)-L(f,P)&=\sum_{k=1}^{n}(M_k-m_k)(x_k-x_{k-1})=\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})]\Delta x\\&=[f(a)-f(b)]\Delta x. \end{align} Since $\lim_{n\to\infty}\Delta x=0$, it follows that $U(f,P)-L(f,P)\to0$. Hence, $f$ is integrable.

Does this seem reasonable? I believe I heard that $\epsilon$ definitions were necessary for this.

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You don't have that $U(f,P)-L(f,P)$ is equal to $0$. You have that the limit as $n\to\infty$ of $U(f,P)-L(f,P)$ is $0$. To prove that $U(f,P)-L(f,P)\lt \epsilon$ for sufficiently large $n$ you would use $\epsilon$ arguments. – Arturo Magidin Apr 24 '12 at 17:20
For a fixed $\epsilon$, you need $n$ large enough so that $[f(b)-f(a)]\Delta x<\epsilon$, that is, you need $n>[f(b)-f(a)](b-a)/\epsilon$. – M Turgeon Apr 24 '12 at 17:25
A continuous function defined on an interval $[a,b]$ is integrable on $[a,b]$. Yours is just a particular case... – Beni Bogosel Apr 24 '12 at 17:38
@Beni Then a proof of the general case will be very welcome here! – Pedro Tamaroff Apr 24 '12 at 17:58
No one's said it explicitly yet, so: any increasing function on $[a,b]$ is integrable. Any continuous function on $[a,b]$ is integrable. Requiring $f$ to be both increasing and continuous is rather excessive, and I don't think it even gives you an easier proof. – Chris Eagle Apr 24 '12 at 21:18

Suppose $f$ is increasing an obvious modification of the argument works for the case when f is decreasing. Let $\epsilon>0$, choose a partition $\pi=(x_k)_{k=0}^{n}$ and define $\mu(\pi)=max\{x_k-x_{k-1}:k=1,2,\dots,n\}$ the partition is such that $|f(b)-f(a)|\mu(\pi)<\epsilon$ For instance, one can choose a positive integer n such that $n>[f(b)-f(a)+1](b-a)/\epsilon$ and define the partition $\pi=(x_k)_{k=0}^{n}$ by $x_k=a+(k/n)(b-a)$ so $x_k-x_{k-1}=(b-a)/n$ for each k, $1\le k\le n$ then $m_k=inf\{f(t):x_{k-1}\le t \le x_k\}=f(x_{k-1})$ and $M_k=sup\{f(t)x_{k-1}\le t \le x_k\}=f(x_k)$, so $$U(\pi,f)-L(\pi,f)=\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})](x_k-x_{k-1})\le\mu(\pi)\sum_{k=1}^{n}[f(x_k)-f(x_{k-1})]=\mu(\pi)[f(b)-f(a)]<\epsilon$$ so f is riemann integrable, Now if f is continuous on $[a,b]$ then is uniformly continuous so there exists $\delta>0$ such that $f(x)-f(y)<\epsilon/(b-a)$ when ever $|x-y|<\delta$ I hope now from the way I did it for increasing function you can prove it for the continuous $f$ which is more easy to show.