Prove that for each e there is a partitionPe of [0,1] such thatU(f, Pe)≤1/2+e.

Question

Let $f:[0,1]\rightarrow \mathbb{R}$ defined by $f(x)= x$, $0\le x \le 1$. Prove that for each $\epsilon > 0$ there is a partition $P$ of [0,1] such that $U(f, P)≤1/2+\epsilon$.

I know I need a partition such that the area of the rectangles that's over the line is very very small. other than that I am lost.

Answer 1

Let $0 < \delta = \frac{1}{n}$, $P = (k\delta)_{0 \le k\le n}$. $$U(f,P) = \sum_{k=0}^{n-1} (k+1)\delta*\delta = \frac{n(n+1)}{2} \delta^2 = \frac{1}{2} + \frac{1}{2n} \le 1/2 + \epsilon $$ if $n \ge \frac{1}{2\epsilon}$