Real Analysis homework help regarding integration proof. I need some help here, folks...
Problem: If $f$ and $g$ are integrable on $[a,b]$ and $g(x) \leq f(x)$ for all $x \in [a,b]$, then $\int_a^b g \leq \int_a^b f$.
Any help would be great! Thanks!
 A: I'll assume it is a riemann integral. Let $f \geq 0$. Observe that the Riemann upper sum and lower sum for this is $\geq 0$. (From Rudin), we define
$$\underline{\int_a^b} f dx = \sup_{\mathbb{P}} \sum m_i \Delta x_i$$
$$\overline{\int_a^b} f dx = \inf_{\mathbb{P}} \sum M_i \Delta x_i$$
where $m_i$ and $M_i$  are the infimum and supremum respectively of the intervals decided by the partitions. Please read the Riemann integrals chapter from Rudin to understand what I am saying. My point is that both the integrals mentioned above are non -ve and so the riemann integral which is equal to both as it exists is non negative. Now substitute f with f-g, use linearity and you get your result.
A: Keeping technical details out of this because it is a homework problem, but please ask where you would need clarification.
Assuming ordinary Riemann Integration, decompose the $\int f$ into a sum of rectangles over a partition. Every $f$-rectangle must be at least as high as the corresponding $g$-rectangle, hence the same limit for $g$ would converge to a smaller value.
