Question: (1) Let's prove that there exist functions with infinitely many discontinuities which are (surprisingly)integrable.Let f:[0,1]Rbe defined byf(x)={1ifx=1mforsomeminN0otherwiseNotice that f has infinitely many discontinuities in this

(1) Let's prove that there exist functions with infinitely many discontinuities which are (surprisingly)integrable.Let f:[0,1]Rbe defined byf(x)={1ifx=1mforsomeminN0otherwiseNotice that f has infinitely many discontinuities in this interval.For NinN, set PNtobe the following partition of0,1 :{0,(1N+12N2),(1N-1-12N2),(1N-1+12N2),dots,(13-12N2),(13+12N2),(12-12N2),(12+12N2),(1-12N2),1}(a) For an integer k with 2kN, prove that every subinterval of the form(1k+12N2),(1k-1-12N2)contains no real numbers of the form x=1m, where misan integer.(b) Prove all the other types of subintervals; the ones of the form[(1k-12N2),(1k+12N2)],where 2kN-1as well as0,(1N+12N2)and(1-12N2),1do contain a real number of the form 1m for minN, as well as real numbers which are not ofthat form.(c) Use (a),(b)to compute U(f,PN) and L(f,PN).(d) Use the integrability criterion, together with help from (c),to prove that fis integrable on0,1.(e) Find 01f.

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Mathematics Questions!