Q:

Why is $L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(P,f)+U(P,g)$

Let $R \subset \Bbb{R}^n$ and $f,g$ be two integrable functions from $R\to \Bbb{R}$ , in a proof of a property of integrals Rudin writes the following: $$L(P,f)+L(P,g) \le L(P,f+g) \le U(P,f+g) \le U(P,f)+U(P,g)$$ where $P$ is any partition. The $\le$ between  $L(P,f+g)$  and $U(P,f+g)$ is clear but why is it that  $L(P,f)+L(P,g) \le L(P,f+g)$  and $U(P,f+g) \le U(P,f)+U(P,g)$ isn't it easy to see by expanding the sums to see that they are equal since $\sup(f+g)=\sup(f)+\sup(g)$ (and similarly for the $\inf$) ? Am I missing something or did Rudin just write it this way since it won't make a difference? 
Note: Rudin's proof is over an interval, but the multivariable case is similar.

A:

You have made a small error. It is true that $\sup(f+g) \leq \sup f + \sup g$. The inequality may be strict though. (Can you come up with an example for yourself? Consider $f$ and $g$ which obtain their maxima at different places.)

