Muddiest Point 11/30

I thought that the muddiest point from last Thursday’s class was when we went over proving that \int kf = k\int f. 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 k>0. Then we see that U(kf,P) =\sum sup\{kf(x): x \in [x_{k-1},x_k] \}. We already know from earlier this semester that csup(A) = sup(cA) 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 U(kf,P) = kU(f,P). we can use the same logic for the lower sums as well. Since L(kf,P) = inf\{kf(x): x \in [x_{k-1},x_k] \} we can once again use our knowledge about infinums and supremums from earlier in the semester that kinf(A) = inf(kA) to justify the statement that inf\{kf(x): x \in [x_{k-1},x_k] \} = kinf\{f(x): x \in [x_{k-1},x_k] \} and L(kf,P) = kL(f,P). Since the lower sums and Upper sums are unchanged by the location of the k we can conclude that if k is positive then \int kf = k\int f and that there exists a partition P_n 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 k is negative. If k is negative U(f,p) will be flipped when multiplied by k and in turn L(f,P) will be flipped when multiplied by k. 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 k we now have that U(kf,P) = kL(f,P) and kU(f,P) = L(kf,P). 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 \int kf = k\int f regardless of the value of k.

One Response

  1. Jeremy LeCrone says:

    Thank you for the post. Good identification of a an issue that really is easy to overlook when you first consider this property.

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