5-16

Suppose that f \ge 0and is measurable on E and that \omega is finite on (0,\infty). Show that Theorem 5.51 holds without any further restrictions (i.e. we don't need to assume f\in L^p(E) and we don't need to assume |E| is finite) if we interpret \int_0^{\infty} \alpha^{p-1}\omega(\alpha) d\alpha=\lim_{a\to 0+, b\to 0+\infty} \int_a^b \alpha^p d \omega(\alpha ). 
Answer: Given that f \ge 0 and |E| is assumed not necessarily finite. |E|could be finite of infinite. 
And from theorem 5.46, if \phi(x)=|x|^ p , then we have \int_{E_{ab}}\phi(f) = - \int_a^b \phi(\alpha)d \omega(\alpha) is finite. 
We write down f|_{E _ {ab}} as the increasing limit of simple measurable functions over partitions of [a,b], and the norms of partitions tends to be 0 (This can be shown in theorem 4.13 (textbook page 68) , Every function f can be written as the limit of a sequence { f_k } of simple functions.). 
And for the Riemman-Stietljes form, \phi is monotone increasing on nonnegative real numbers, and we have \varphi\left(f_{k}\right) \nearrow \varphi(f), so by the Monotone Covergence theorem, we obtain 
\int_{E_{a b}} \varphi\left(f_{k}\right) \rightarrow \int_{E_{a b}} \varphi(f). 
\varphi(f _ k) is a sequence of simple functions , we can get 
\sum_{j} \varphi\left(\alpha_{j-1}^{(k)}\right)\left[\omega\left(\alpha_{j}^{(k)}\right)-\omega\left(\alpha_{j-1}^{(k)}\right)\right] \leq \int_{E_{a b}} \varphi\left(f_{k}\right) \leq \sum_{j} \varphi\left(\alpha_{j}^{(k)}\right)\left[\omega\left(\alpha_{j}^{(k)}\right)-\omega\left(\alpha_{j-1}^{(k)}\right)\right]. 
where \alpha_j^{(k)} is the j th partition element of [a.b] for the f_k The norms of partitions approaches 0 as k \rightarrow \infty, and \int_{E_{a b}} \varphi\left(f_{k}\right) \rightarrow-\int_{a}^{b} \varphi(\alpha) d \omega(\alpha), we let a \rightarrow 0 and b \rightarrow +\infty, then by monotone covergence theorem it shows that 
\int_{E} \varphi\left(f_{k}\right) \rightarrow-\int_{0}^{ \infty} \varphi(\alpha) d \omega(\alpha).
Case1 (FINITE CASE) : Suppose that \int _0 ^ \infty \alpha ^ p d \omega ( \alpha ) is finite, then f \in L^ p ( E ), we using the integrate by parts and we can get
\int_{a}^{b} \alpha^{p} d \omega(\alpha)=\left.\alpha^{p} \omega(\alpha)\right|_{a} ^{b}-p \int_{a}^{b} \alpha^{p-1} \omega(\alpha) d \alpha \rightarrow- p \int_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d \alpha 
Conversely, \int _0 ^ \infty \alpha ^ {p-1} d \omega ( \alpha ) is finite, and integrating by parts again, and {lim}_{ a \rightarrow 0 } a ^p \omega(a) equal to 0 by exercise 15. 
\left.\alpha^{p} \omega(\alpha)\right|_{a} ^{b}-p \int_{a}^{b} \alpha^{p-1} \omega(\alpha) d \alpha= \int_{a}^{b} \alpha^{p} d \omega(\alpha) .
As a \rightarrow 0 and b \rightarrow +\infty in both sides all get what we want, \int_{E} f^{p}=-\int_{0}^{\infty} \alpha^{p} d \omega(\alpha)=p \int_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d\alpha
Case II, if \int _0 ^ \infty \alpha ^ p d \omega ( \alpha ) or \int _0 ^ \infty \alpha ^ {p-1} d \omega ( \alpha ) is not finite, then they are both equal to + \infty , as f is nonnegative in our assuption. Obviously \int_{E} f^{p}=-\int_{0}^{\infty} \alpha^{p} d \omega(\alpha)=p \int_{0}^{\infty} \alpha^{p-1} \omega(\alpha) d\alpha = + \infty