1.6. Differentiation theorems 139 function on R, and F vanishes outside of [a, b]. As F is almost everywhere differentiable, the Newton quotients fn(x) := F (x + 1/n) − F (x) 1/n converge pointwise almost everywhere to F . Applying Fatou’s lemma (Corollary 1.4.46), we conclude that [a,b] F (x) dx ≤ lim inf n→∞ [a,b] F (x + 1/n) − F (x) 1/n dx. The right-hand side can be rearranged as lim inf n→∞ n [a+1/n,b+1/n] F (y) dy − [a,b] F (x) dx which can be rearranged further as lim inf n→∞ n [b,b+1/n] F (x) dx − [a,a+1/n] F (x) dx . Since F is equal to F (b) for the first integral and is at least F (a) for the second integral, this expression is at most ≤ lim inf n→∞ n(F (b)/n − F (a)/n) = F (b) − F (a) and the claim follows. Exercise 1.6.44. Show that any function of bounded variation has an (al- most everywhere defined) derivative that is absolutely integrable. In the Lipschitz case, one can do better: Exercise 1.6.45 (Second fundamental theorem for Lipschitz functions). Let F : [a, b] → R be Lipschitz continuous. Show that [a,b] F (x) dx = F (b) − F (a). (Hint: Argue as in the proof of Proposition 1.6.37, but use the dominated convergence theorem (Theorem 1.4.48) in place of Fatou’s lemma (Corollary 1.4.46).) Exercise 1.6.46 (Integration by parts formula). Let F, G: [a, b] → R be Lipschitz continuous functions. Show that [a,b] F (x)G(x) dx = F (b)G(b) − F (a)G(a) − [a,b] F (x)G (x) dx. (Hint: First show that the product of two Lipschitz continuous functions on [a, b] is again Lipschitz continuous.)
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