prove $\int f \le \int g$ After Edwards(author of advanced calculus of several variables) pushed me through contented sets and admissible functions, he threw me a curveball. I have a pretty good idea on how to prove this with the two excluded properties... but without... well any hints/solutions would be appreciated: If $f,g$ are integrable functions with $f(x) \le g(x)$ for all $x$, prove $\int fdx \le \int gdx$ without using the fact that a set of integrable functions is a vector space and the mapping $\int:Q \rightarrow R$ is linear (with Q being the set of integrable functions) Thanks guys!

According to the above comments the integral being used is Riemann which is equivalent to the Darboux integral hence we may use Upper and Lower sums to prove this. Since we assume the functions f and g to both be Darboux integrable then we may use only the Upper sums; Let $\displaystyle\sum_{i=1}^{n}M_{i}^{f}(t_i-t_{i-1})$ be the upper sum for f and $\displaystyle\sum_{i=1}^{n}M_{i}^{g}(t_i-t_{i-1})$ be the upper sum for g.

Given some partition of $[a,b]$ then note that $f(x)\le g(x)\le M_{i}^{g}$ for all $x\in [t_{i-1} , t_i]$ and i=1,...,n. Hence $M_{i}^{f}\le M_{i}^{g}$ for each i=1,...,n. So

$\displaystyle\int_{a}^{b}f(x)dx=\inf_{\text{partitions P of }[a,b]}\bigg(\sum_{i=1}^{n}M_{i}^{f}(t_i-t_{i-1})\bigg)\le \sum_{i=1}^{n}M_{i}^{g}(t_i-t_{i-1})$.

This is true for all partitions so

$\displaystyle\int_{a}^{b}f(x)dx\le\inf_{\text{partitions P of }[a,b]}\bigg(\sum_{i=1}^{n}M_{i}^{g}(t_i-t_{i-1})\bigg)=\int_{a}^{b}g(x)dx$ as we wanted.