I thought that the muddiest point from last Thursday’s class was when we went over proving that . The reason why I thought that this was the muddiest is because the proof required to cases. A case in which k was negative and a case that case was positive . This is often a detail that is easy to look over as it seems so trivial which is why it is important to go over it. To start the proof we assume that f is integrable and that . Then we see that . We already know from earlier this semester that therefore we can say that $latex sup\{kf(x): x \in [x_{k-1},x_k] \} = ksup\{f(x): x \in [x_{k-1},x_k] \}$ and in turn . we can use the same logic for the lower sums as well. Since  we can once again use our knowledge about infinums and supremums from earlier in the semester that to justify the statement that  and . Since the lower sums and Upper sums are unchanged by the location of the we can conclude that if is positive then  and that there exists a partition such that $ \lim_{n \to \infty} [U(kf, P_n) − L(kf, P_n)] = \lim_{n \to \infty} k[U(f,P_n) − L(f,P_n)] = 0$. Now we must consider the case that is negative. If k is negative will be flipped when multiplied by k and in turn will be flipped when multiplied by . This distinction is important to make and is why the second case is needed. With the lower and upper sums flipped by the multiplication by we now have that and . Therefore it is still the case that $ \lim_{n \rightarrow \infty} [U(kf, P_n) − L(kf, P_n)] = \lim_{n \rightarrow \infty} k[U(f,P_n) − L(f,P_n)] = 0$. So we can conclude that  regardless of the value of .