Let $Q$ be a rectangle in $E^n$ and assume that $f: Q\longrightarrow E$ is integrable. Show that if $f(x) \geq 0$, for all $x \in Q$, then $\int Q f ≥ 0.$
Solution:
For every partition $P$ of $Q$, its lower sum $L(f, P)$ is non-negative, so $\int Q f\geq L(f, P) \geq 0$.
I do not understand the solution fully is it missing a step using the definition $\lim_{\text{mesh}\to0}\sum f(_)\times(x_{i+1}−x_i)$? I do not know how to use the definition here.
an integral is always larger then it's lower sum on any partition, that is what the proof is based on. If you want to prove it using the limit, you can show that each element in the sequence is none negative and therefore the limit is none negative.