1.3. The Lebesgue integral 59 for instance, the absolutely integrable analogue of Exercise 1.3.17 tells us that b a f(x) dx = [a,b] f(x) dx for any Riemann-integrable f : [a, b] C. Exercise 1.3.22. If E, F are disjoint measurable subsets of Rd, and f : E F C is absolutely integrable, show that E f(x) dx = E∪F f(x)1E(x) dx and E f(x) dx + F f(x) dx = E∪F f(x) dx. We will study the properties of the absolutely convergent Lebesgue in- tegral in more detail in later notes, as a special case of the more general Lebesgue integration theory on abstract measure spaces. For now, we record one very basic inequality: Lemma 1.3.19 (Triangle inequality). Let f L1(Rd C). Then | Rd f(x) dx| Rd |f(x)| dx. Proof. If f is real-valued, then |f| = f+ + f− and the claim is obvious from (1.12). When f is complex-valued, one cannot argue quite so simply a naive mimicking of the real-valued argument would lose a factor of 2, giving the inferior bound | Rd f(x) dx| 2 Rd |f(x)| dx. To do better, we exploit the phase rotation invariance properties of the absolute value operation and of the integral, as follows. Note that for any complex number z, one can write |z| as zeiθ for some real θ. In particular, we have | Rd f(x) dx| = eiθ Rd f(x) dx = Rd eiθf(x) dx
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