- #1

bonfire09

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## Homework Statement

Let ##f:[a,b]\rightarrow\mathbb{R}## and ##g:[a,b]\rightarrow\mathbb{R}## be continuous functions having the property ##f(x)\leq g(x)## for all ##x\in[a,b]##. Prove ##\int_a^b \mathrm f <\int_a^b\mathrm g## iff there exists a point ##x_0## in ##[a,b]## at which ##f(x_0)<g(x_0)##. I just needed help with the reverse direction.

## Homework Equations

##R(f,P_n)## stands for the Riemann integral.

## The Attempt at a Solution

Suppose there exists a point ##x_0## in ##[a,b]## such that ##f(x_0)<g(x_0)##. Let ##P_n## be a regular partition for ##[a,b]## such that ##\lim_{n\to\infty} ||P_n||=0##. Then there exists an interval ##[x_{j-1},x_j]## such that ##x_0\in[x_{j-1},x_j]##. Let ##c_j=x_0## for some tag ##c_j## where ##1\leq j\leq n##. Picking the same tags for both functions we have ##R(f,P_n)=\sum_{i=1}^{n} f(c_i)(x_i-x_{i-1})=f(c_1)(x_1-x_0)+...+ f(c_j)(x_j-x_{j-1})+...+f(c_n)(x_n-x_{n-1})## and ##R(g,P_n)= \sum_{i=1}^{n} g(c_i)(x_i-x_{i-1})=g(c_1)(x_1-x_0)+...+g(c_j)(x_j-x_{j-1})+..._g(c_n)(x_n-x_{n-1})##. We see by assumption that ##f(c_j)(x_j-x_{j-1})< g(c_j)(x_j-x_{j-1})##. Then ##f(c_1)(x_1-x_0)+...+ f(c_j)(x_j-x_{j-1})+...+f(c_n)(x_n-x_{n-1})< g(c_1)(x_1-x_0)+...+g(c_j)(x_j-x_{j-1})+..._g(c_n)(x_n-x_{n-1})##. Thus ##R(f,P_n)=\sum_{i=1}^{n} f(c_i)(x_i-x_{i-1})<R(g,P_n)= \sum_{i=1}^{n} g(c_i)(x_i-x_{i-1}) \implies R(f,P_n)<R(g,P_n)##. Since ##f## and ##g## are continuous that means they are integrable. Thus we have ##\int_a^b \mathrm f =\lim_{n\to\infty} R(f,P_n)<\lim_{n\to\infty} R(g,P_n)=\int_a^b\mathrm g##

Here is my solution. I need to know if its right or not and how I can clean it up. Thanks.

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