# Math Help - Proof of Riemann integrable function

1. ## Proof of Riemann integrable function

Let $f : [a,b] \rightarrow \mathbb{R}$ be increasing on the interval $[a,b]$ (that is, $f(x) \leq f(y)$ whenever $x < y$, so in other words monotone increasing). Prove $f$ is integrable on $[a,b]$.

By integrable, we're referring to Riemann integrable.

Any way, uhh help? Would I go about proving this saying that if we can prove its continuous, then its integrable? That's some theorem. But isn't it obvious it's continuous, as it says its increasing

2. Originally Posted by Auxiliary
Let $f : [a,b] \rightarrow \mathbb{R}$ be increasing on the interval $[a,b]$ (that is, $f(x) \leq f(y)$ whenever $x < y$, so in other words monotone increasing). Prove $f$ is integrable on $[a,b]$.
Note $f(a)\leq f(x)\leq f(b)$ thus it is a bounded function.
We need to show for any $\epsilon > 0$ we can choose a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)$. Choose a partition $P = \{x_0,x_1,...,x_n\}$ such that $\delta = \max_{1\leq k\leq n} (x_k - x_{k-1}) < \frac{\epsilon}{f(b)-f(a)}$. Then $U(f,P) - L(f,P) = \sum_{k=1}^n [\sup\{ f(x): [x_{k-1},x_k]\} - \inf\{ f(x): [x_{k-1},x_k]\}] (x_k-x_{k-1})$ $\leq \sum_{k=1}^n [f(x_k)-f(x_{k-1})]\delta = (f(b)-f(a))\delta < \epsilon$