1. ## riemann integrable

Let [0,1] ---> R be bounded. Let f be decreasing. By consideration of the sequence of dissection (Dn) of [0,1] show that f is integrable.

Let E > 0. Then for all N such that s(Dw) < E.

Hence it satisfies Riemann's criteria for integrability.

Can some please help me with this question, as my answer as it stands is rubbish and obviously too short, but I don't see what else to say

2. Originally Posted by hunkydory19
Let [0,1] ---> R be bounded. Let f be decreasing. By consideration of the sequence of dissection (Dn) of [0,1] show that f is integrable.

Let E > 0. Then for all N such that s(Dw) < E.

Hence it satisfies Riemann's criteria for integrability.

Can some please help me with this question, as my answer as it stands is rubbish and obviously too short, but I don't see what else to say

Suppose that f is increasing on [0,1] then if $0\leq x\leq 1$ we have $f(0)\leq f(x)\leq f(1)$ thus f is a bounded function. Let P be a partition of [0,1] then difference between the upper and lower sum is $\sum_{k=1}^n(\sup\{ f:[x_{k-1},x_k]\} - \inf \{ f: [x_{k-1},x_k]\} )(x_k-x_{k-1}) =$ $\sum_{k=1}^n (f(x_k) - f(x_{k-1})(x_k - x_{k-1}) \leq \sum_{k=1}^n (f(x_k) - f(x_{k-1})\delta = (f(1) - f(0))\delta$ where $\delta = \max_{1\leq k\leq n} (x_k - x_{k-1})$. Thus, if we choose $\epsilon < \frac{\delta}{f(1)-f(0)}$ then we insure that the upper and lower sums are withing $\epsilon$ of eachother. And so the function is integrable.