Answer to Question #170730 in Real Analysis for jan

Question

Prove the following result:

A function f that is decreesing on [a,b] is integrable on [a,b].

Answer 1

Given that f is decreasing on [a,b]. Thus, $f(b) \leq f(x) \leq f(a)$ and f is bounded on [a,b]. Given $\epsilon >0 \, \, \exist k >0 \ni k[f(a)-f(b)]< \epsilon$ .

Let P = $\{x_0, x_1,..., x_n\} \ni ∆x_i \leq k$ be a partition on [a,b].

Since f is decreasing it follows that $m_i = f(x_i)$ and $M_i = f(x_{i-1}) i =1,2,...,n$

where $m_i$ is the greatest lower bound of f on $[x_{i-1},x_i]$ and $M_i$ is the lowest upper bound of f on the interval $[x_{i-1},x_i]$ $U(f,P) - L(f,P) = \sum_{i=1}^{n}[f(x_{i-1})-f(x_i)]∆x_i \leq k\sum_{i=1}^{n}[f(x_{i-1})-f(x_i)]$ = $k[f(x_0)-f(x_n)]=k[f(a)-f(b)]< \epsilon$

Hence f is integrable on [a,b].