Let be $f$ and $cf$ integrable functions on the interval $[a,b ]$ where $c$ is a constant with $c<0$.
Show that: $$\int_a^b cf = c \int_a^b f.$$ I know that there is a different approach where you directly use $\sup$ and $\inf$. However, I want to know if the following approach is correct:
We already know from lecture that there exists a partition $P$ such that $U(f, P)- L(f, P) < \frac{\epsilon}{-c}$, where $U, L$ are the corresponding Darboux-sums and $\epsilon>0$.
Rearranging yields: $$cL(f, P)-cU(f,P)= U(cf,P)-cU(f,P) < \epsilon.$$ This inequality still holds with increasing refinement of $P$. Hence, taking the $\sup$ of both Darboux-sums we get $\int_a^b cf - c \int_a^b f <\epsilon$ as both functions $f$ and $cf$ are integrable on the interval $[a,b ]$. This is equivalent to $\int_a^b cf = c \int_a^b f$.
Is this correct?