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Unformatted text preview: Real Analysis - Math 630
Homework Set #6 - Chapter 4
by Bobby Rohde 10-12-00 Problem 10 a) Show that if f is integrable over E then so is f and
Ef E f Does the integrability of f imply integrability of f? Proof
f is integrable f and f are each integrable. However f = f + f , so by Proposition 15ii, we have that f is integrable and Also Ef Ef Ef . Ef Ef Ef Ef E f , but since f and f are nonnegative we can drop the asbsolute value signs, so E f Ef Ef E f . So we have shown the first part. f f f , So f integrable ( f + f ) is integrable. However the set A xfx 0 is disjoint from the set B xfx 0 , and A and B must be measurable iff f is a measurable function. Thus by Proposition 15, E f f f ABf Af B f . Thus f and f are each integrable iff f is a measurable function. But this implies that A f Bf ABf integrability of f iff f is a measurable function. f
Ef , is integrable. Thus f implies MATH630-6.nb 2 b) The improper Riemann integral of a function may exist without the function being integrable. If f is integrable, show that the improper Riemann integral is equal to the Lebesgue integral whenever the former exists. Proof
Let R a f x, be a Riemann integrable function with improper limit point a, and f defined for all x (a, b]. Note that if any other point in (a, b] is improper, we may into two or more integrals each of which with one improper limit point so we need only assume one limit is improper. In particular we know from analysis that improper points must be disconnected in order for a function to be Riemann integrable. WLOG I assume that a < b, but all parts of the proof may be reversed for a > b, and will still hold accordingly. Case |a| < Then consider the sequence of measurable integrable functions f x, a 1 n x 0, otherwise Theorem 16 we have that fn x lim
b a fn b f n , such that f . Thus by and hence b
b af . Then lim fn x = f x a.e. Also we have that fn lim R
b a fn . b a f. b a fn But by construction we must have that fn x is So
b af bounded, which allows us to apply Proposition 4 to get that lim R
b a fn R b a lim fn R b a f. =R b a fn , Case a = Then consider the sequence of measurable integrable functions fn x f n , such that f x, n x b . Then lim fn x = f x a.e. Also we have that fn f . Thus by 0, otherwise b b Theorem 16 we have that a f lim a fn . But by construction we must have that fn x is bounded, which allows us to apply Proposition 4 to get that lim
b a fn lim R b a fn R b a lim fn R b a f. So b af R b a f. b a fn =R b a fn , and hence QED MATH630-6.nb 3 Problem 11 If is a simple function, we have two definitions for , n and i 1 ai m Ai . Show that they are equal. Proof
Consider , By problem 4b, we know that nonnegative measurable functions f, f sup , with the sup taken over all simple functions f, and the right hand side evaluated according to the old rule for integrating step functions. Since it follows immediately from the definition of f and f that and are simple, we can thus deduce that n1 n2 n1 sup sup i 1 bi mBi , i 1 ci mCi , where i 1 bi Bi and n2 n1 n2 So i 1 ci Ci . i 1 bi mBi i 1 ci mCi , also we know that Cj i, j and each bi corresponds to some a j 0 and each ci corresponds to some Bi n1 n2 n1 n2 n aj 0, so i 1 bi mBi ci mCi i 1 ai mAi . Since the i 1 ci mCi i 1 bi mBi i1 sets Ci and Bi , must follow the same correspondence. QED. Problem 12 Let g be an integrable function on a set E and suppose that f n is a sequence of measurable functions such that fn x g x a.e. on E. Then
E lim fn
E fn E lim
lim E fn lim E fn E lim fn Proof
We know for free that lim We only need to show that
E fn , since this is true for any lim sups and infs.
E fn lim fn lim and lim E fn E lim fn . fn is measurable and bounded by an integrable function, therefore fn is integrable. Also we know that lim fn fn n thus By Proposition 15iii, we have that E lim fn n E fn lim E fn . A similar argument for lim sup fn fn gives us that lim E fn E lim fn E lim fn , and thus we are done. QED MATH630-6.nb 4 Problem 15 Let f be integrable over E. Then given > 0,
a) a simple function
E such that f Construction Since E f , it suffices to show that the Ef Ef E Ef Ef E latter is less than . If we further specify that A xfx 0 and B xfx 0 , then we may clearly choose = 0, x E A B . So we may further simplify the above expression to E f , Ef E Ef Ef A B Ef A Ef B which again all that we need to show is that this new expression is less than . By Problem 4, we know that nonnegative measurable functions f we have that f sup , taken over all simple functions f. Since f and f are nonnegative measurable functions we know there must exist simple functions S and I respectively such that 0 f 2 S and 0 f 2. The second implies that 0 f 2. Thus if we I I choose over A = and over B = then we have that S I, 2 2 . Thus we have shown the construction as Ef A Ef B required. a) a step function
E such that f Construction By the Question 3.23c, which part of the proof to Proposition 22, we know that a step function such that = a simple function except on a set of less than / 3 and if M bounds then M bounds , > 0. Let > 0 and be a step function such that E f 2. Let N be a number such that x N , x. This is possible since simple functions are always bounded. Choose such that = except on a set of measure 4 N with the set denoted by A, and bounded by N. Thus < EA f + = EA f + Ef EAf Af Af +Af + Af +A Af EA f EA f / 2 + 2*N *mA = / 2 + 2*N* 4 N = . Therefore
E f . MATH630-6.nb 5 a) a continuous function g such that
E f g Construction By question 3.23d, we have that a continuous function g such that g = a step function except on a set of measure less than / 3, which agress with the bounds on . With this we may replace the occurances of in the above proof with and with g and the proof will precede identically, except that we have to infer the boundedness of from the boundedness of the preceding . Thus we may construct g such that E f g . ...
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This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.
- Spring '08