5-7

Question: Give an example of an f that is not integrable, but whose improper Riemann integral exists and is finite. 
Solution: 
Define a function f: E\to \mathbb{R} where E = [0,\infty] is measurable such that f(x)=\frac{\sin{x}}{x} is a measurable function. 
Then we have \int_0^\infty \frac{\sin{x}}{x} \,dx = \lim_{t\to\infty} \int_0^t \frac{\sin{x}}{x} \,dx = \frac{\pi}{2} . Hence there exists as an improper Riemann integral such as \int_0^\infty \frac{\sin{x}}{x} \,dx = \frac{\pi}{2} , which is finite. 
From Theorem 5.21, since \int_0^\infty \left|\frac{\sin{x}}{x}\right| \,dx \notin L(E) , which is not finite, so it is not integrable. This implies \int_0^\infty \frac{\sin{x}}{x} \,dx \notin L(E) by Theorem 5.21. 
(Note: Theorem 5.21 says, let f be measurable on E. Then f is integrable over E iff |f| is. )