Riemann Integrals, etc.Date: 24 Apr 1995 23:12:25 -0400 From: Anonymous Subject: first of three problems Hi ..this has something to do with the Riemann Integral..please give me a hint on how to start this! Let A be a subset of the Reals, Prove that: i) If A is finite, then the outer measure is zero ii) If A = [a,b], then the outer measure = (b - a) Next problem: Let f and g be elements of the Reals, and let I be a subset of the Reals. { x is an element of I such that f(x) does not equal g(x)} is finite Then the integral of f, from a to b equals the integral of g from a to b Date: 28 Apr 1995 15:23:31 -0400 From: Dr. Ken Subject: Re: first of three problems Hello there! >Let A be a subset of the Reals, Prove that: >i) If A is finite, then the outer measure is zero The key thing in showing that the outer measure for some set is zero is to show that the measure is less than every possible positive number. Let e (epsilon) be a small positive number. See whether you can show that there is some covering of A that has measure less than e. Hint: if there are n points in the set, how big can the covering around each point be so that the total covering is less than e? >ii) If A = [a,b], then the outer measure = (b - a) This is a similar problem, and I think it will be easier to do once you've done the first part. If not, write back and we'll try to help you out. >let f and g be elements of the Reals ,and let I be a subset of the >Reals. > > { x is an element of I such that f(x) does not equal g(x)} is finite > >Then the integral of f, from a to b equals the integral of g > from a to b You (hopefully) showed that a finite set of points in the real number has outer measure zero. There's a theorem that says that if two functions only disagree on a set of measure zero in an interval, then their integrals on that interval are equal. To prove that, you could break up the integrals into lots of smaller integrals, where some of the integrals have only points where f and g agree, and some of the integrals are centered around points where f and g disagree. Then show how you can make the total area found with the second kind of integrals really small. -Ken "Dr." Math Date: 29 Apr 1995 11:18:52 -0400 From: Anonymous Subject: Re: first of three problems Hi Thanks for the hints, I'll do that and get back to you.. Here's the last problem Suppose (f sub n)n converges uniformly to f. Then, If each function f sub n is bounded, so is f Thanks Mike Date: 30 Apr 1995 12:38:36 -0400 From: Dr. Sydney Subject: Re: first of three problems Dear Mike, Hi! Good question. I'll try to give you a few ideas for directions in which you might go, and maybe you can take it from there... Write down the explicit definitions of uniform convergence of f(sub n) to f and bounded f... Now maybe you can work with a triangle inequality in which f (sub n) and f are involved. |f(x)| = |f(x) - f(sub n)(x) + f(subn)(x)|, right? Now make this into a triangle inequality, and then apply the definition of uniform convergence for a specific epsilon, say epsilon = 1/2. If these hints are too vague or are confusing, or if you want to check your answer just let us know! Good luck! --Sydney, "Dr. Math" |
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