Question

Please answer it step by step and Question 2. uniformly converge is defined by *f=0* clear handwritten, please, also, beware that for the x you have 2 conditions , such as x>n and 0<=x<=n

1- For all n > 1 define fn: [0, 1] → R as follows: (i if n!x is an integer 10 otherwise Prove that fn + f pointwise where f:[0,1] → R is defined by ſo if x is irrational f(x) = 3 11 if x is rational Hint: • If c € (0,1) is irrational, then for all n, fn(c)=0. • If c € (0,1) is a rational number, then c= ? for some nonnegative integers p and q. Clearly for all n >q, n!c is an integers and so for all n greater than 4, fu(c)=1. 2- Recall that in class we stated that if a sequence of functions (in) converges uniformly on (a,b) to a function f, and if each fn is Riemann integrable on [a,b], then f is Riemann integrable on (a, b) and lim ["fndx = 1° limfndx. noo Ja Ja noo In this exercise you will show that for improper Riemann integrals, even uniform convergence does not necessarily do the job! Recall that if g: 0,00) + R is Riemann integrable on the closed and bounded interval [0, R) for all R > 0, then the improper R Riemann integral $(x)dx is defined as a limit: 8(x)dx = lim (“s(x)dx. RtoJo 2-1) For all n > 1 define fn: [0,) + R as follows: if 0<x<n fuct) = {ő if x>n Prove that fn + f uniformly where f:[0,-) +R is defined by f = 0. 2-2) Show that lim ſº fmdx+ Sºfdx.

Answer 1

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