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Math Forum / Mathematics / General Topics / February 2006Riemann-integrable function
Hello! Let a,b be real numbers and f:[a,b]->R a Riemann-integrable function (R is the set of the real numbers). Let [A,B] \subset R and f([a,b]) \subset [A,B]. Show that every continuous function g:[A,B]->R, g o f : [a,b]->R is Riemann-integrable. ( o denotes the composition of the two functions) How do I show that? Regards, Jonas > Let a,b be real numbers and f:[a,b]->R a Riemann-integrable function (R > is the set of the real numbers). [quoted text clipped - 7 lines] > > How do I show that? If you can use the fact that, for a bounded function, Riemann-integrable is equivalent to the fact that the set of points of discontinuity has Lebesgue measure zero, then it's really very easy. Otherwise, it's a bit more complicated. Best regards, Jose Carlos Santos > Hello! > [quoted text clipped - 9 lines] > > How do I show that? Write down precisely what it means for f to be riemann integrable (what limits must converge), write down what it means for g to be continuous in delta/epsilon and further write down precisely what it would mean for "g o f" to be riemann integrable (it will be similar to f, but with "g o f" instead). If you for example use the upper and lower darbaux sums, I'm pretty sure the solution will jump up and bite you. I think it will be very useful to know that g is continuous on a compact interval so it is uniformly continuous and that will allow you to wiggle the limits for "f" to look like the limits for "g o f". Jiri In article <1139485555.609888.268180@z14g2000cwz.googlegroups.com>, > Hello! > [quoted text clipped - 5 lines] > Show that every continuous function g:[A,B]->R, g o f : [a,b]->R is > Riemann-integrable. Products and sums of RI functions are RI. It follows that g o f is RI if g is a polynomial. By Weierstrass, every continuous function on [A,B] is the the uniform limit of polynomials on [A,B]. Because the uniform limit of RI functions is RI, we're done. (This proof may do you no good.) Here is a solution: http://www.math.nus.edu.sg/~matngtb/Calculus/Comp_Riemann/Comp_Riemann.htm at "Another Proof of Theorem 2" |
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