Proving if $f \leq g$ then $\int_A f \leq \int_A g$

59 Views Asked by At

This comes from Spivak "Calculus on Manifolds" problem 3-5. I think I have proof but it seems too simple to be true so I would like others to look through it and maybe point me to some subtle issues.

I want to prove that if $f,g$ are functions $A \to \mathbb{R}$ with $f \leq g$ where $A \subset \mathbb{R}^n$ is a closed rectangle then their respective integrals are:

$$ \int_A f \leq \int_A g.$$

My attempt is that if $f \leq g$ for $A$ then if I consider any subrectangle $S$ of $A$ then $\mathrm{inf}\{f(x): x\textrm{ in }S\} \leq \mathrm{inf}\{g(x): x\textrm{ in }S\} $. But in that case, if I consider any partition $P$ of $A$ I have for the lower sums $L(f, P) \leq L(g,P)$ where $L(f,P)$ means lower sum of function $f$ for partition $P$. Because this is true for all possible partitions $P$, I can conclude that $\mathrm{sup} \{ L(f,P) \} \leq \mathrm{sup} \{ L(g,P) \}$. But this is exactly the statement of the problem.

I would really appreciate your thoughts and comments.