1.6. Differentiation theorems 121 so we see that the first conclusion of Theorem 1.6.19 implies the second. A point x for which (1.26) holds is called a Lebesgue point of f thus, for an absolutely integrable function f, almost every point in Rd will be a Lebesgue point for Rd. Exercise 1.6.14. Call a function f : Rd → C locally integrable if, for every x ∈ Rd, there exists an open neighbourhood of x on which f is absolutely integrable. (i) Show that f is locally integrable if and only if B(0,r) |f(x)| dx ∞ for all r 0. (ii) Show that Theorem 1.6.19 implies a generalisation of itself in which the condition of absolute integrability of f is weakened to local integrability. Exercise 1.6.15. For each h 0, let Eh be a subset of B(0,h) with the property that m(Eh) ≥ cm(B(0,h)) for some c 0 independent of h. Show that if f : Rd → C is locally integrable, and x is a Lebesgue point of f, then lim h→0 1 m(Eh) x+Eh f(y) dy = f(x). Conclude that Theorem 1.6.19 implies Theorem 1.6.12. To prove Theorem 1.6.19, we use the density argument. The dense subclass case is easy: Exercise 1.6.16. Show that Theorem 1.6.19 holds whenever f is continu- ous. The quantitative estimate needed is the following: Theorem 1.6.20 (Hardy-Littlewood maximal inequality). Let f : Rd → C be an absolutely integrable function, and let λ 0. Then m({x ∈ Rd : sup r0 1 m(B(x, r)) B(x,r) |f(y)| dy ≥ λ}) ≤ Cd λ R |f(t)| dt for some constant Cd 0 depending only on d. Remark 1.6.21. The expression supr0 1 m(B(x,r)) B(x,r) |f(y)| dy ≥ λ} is known as the Hardy-Littlewood maximal function of f, and is often denoted Mf(x). It is an important function in the field of (real-variable) harmonic analysis. Exercise 1.6.17. Use the density argument to show that Theorem 1.6.20 implies Theorem 1.6.19.
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