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royden4

# royden4 - Royden Real Analysis 3rd ed Chapter 4 Problem 4-2...

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Royden, Real Analysis 3rd ed. Chapter 4 Problem 4-2 a. Let h be the upper envelop of f . It is known from Problem 2-51 that h ( y ) = max μ f ( y ) , lim x y f ( x ) . (1) If ϕ f is a step function, it is clear from (1) that ϕ ( y ) lim x y f ( x ) h ( y ) (2) at each continuous point y of ϕ . Hence (2) holds except at the jump points of ϕ , which are fi nite. As a result, Z b a h Z b a ϕ and thus R Z b a f Z b a h. (3) By Problem 2-50 g , there exists step functions ϕ n such that ϕ n h. By Bounded Convergence Theorem, R Z b a f lim Z b a ϕ n = lim Z b a ϕ n = Z b a h. (4) The conclusion follows from (3) and (4) . b. Let g be the lower envelop of f. Similar to part a, we have R Z b a f = Z b a g. Hence f is Riemann integrable if and only if Z b a g = Z b a h. (5) Because g f h, (5) implies g ( y ) = h ( y ) a.e. (See Problem 4-3). The conclusion follows from Problem 2-51 a . Problem 4-3 Let A n = { x : f ( x ) > 1 /n } for n = 1 , 2 , . . . . Since { x : f ( x ) 6 = 0 } = { x : f ( x ) > 0 } = n =1 A n , it su ces to show that m ( A n ) = 0 for each n 1 , which is shown as below 0 = Z f Z f · χ A n Z n 1 · χ A n = n 1 m ( A n ) . 1

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Problem 4-4 a. It is clear that f n f. But f n is not simple. Let E n,k = { x : k · 2 n < f ( x ) ( k + 1) · 2 n } . De fi ne g n ( x ) = 2 2 n X k = 2 2 n k · 2 n χ E n,k ( x ) . Then | g n | 2 n and g n f. There is still no guarantee that m { x : g n ( x ) 6 = 0 } < . This can be remedied by introducing ϕ n ( x ) = g n ( x ) · χ [ n,n ] ( x ) . b. By Proposition 8(iii), Z f sup ½Z ϕ : ϕ is a simple function with ϕ f ¾ . Applying the Monotone Convergence Theorem to { ϕ n } in part a. sup Z ϕ lim Z ϕ n = Z f. The conclusion follows. Problem 4-5 Note that F ( x ) = Z x −∞ f ( t ) dt = Z f ( t ) χ ( −∞ ,x ] ( t ) dt. Since χ ( −∞ ,x n ] χ ( −∞ ,x ] everywhere except possibly at x for any x n x, we have from Theorem 10, lim n →∞ F ( x n ) = F ( x ) for any x n x. (1) That is F ( x ) = F ( x ) . In order to show F ( x +) = F ( x ) , it su ces to verify (1) still holds when x n x. Suppose all x n x + 1 . Then by Theorem 10, F ( x + 1) F ( x n ) = Z f ( t ) χ ( x n ,x +1] Z f ( t ) χ ( x,x +1] = F ( x + 1) F ( x ) .
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