Cfrgtkky

0 $begingroup$ " Suppose $f : [0,1] to mathbb{R}$ is given by $f(1/n) = 1/n$ when $n in mathbb{N}$ and $f(x) = 0$ for all other $x in [0,1]$ . Show that for some partition $P$ of $[0,1]$ , we have $U(f,P) < frac{2}{100}$ " . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$ . Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$ . But it turns out not to work well. How can I approach to solve this? (Let $M_k = sup { f(x) : x in [x_{k-1}, x_k ] }$ the upper sum $U(f,P) = sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = { 0 = x_0 < x_1 < ... < x_n = 1 }$ ) real-analysis riemann-integration share | cite | improve this ...