Let f(x) = x sgn(sin(1/x)) if x != 0
f(x) = 0 if x = 0
on the interval I=[-1,1]
Now I 'm asked to show that f(x) is not piecewise continuous on I and later, I must show that f is integrable on I.
The Attempt at a Solution
I am completely lost here and don't know how to proceed...
I think we should first divide the partition in some regular subpartitions so that x=0 is excluded (the only possible discontinuity?):
P1=[-1, -e/100] and P2=[e/100, 1], and then try to show that continuity on the interval P3=[-e/100, e/100] is impossible but that we still have uniform continuity on P1 and P2?
Should I be looking in that direction? Or should I look at it in a complete different way?
Thank you for the help!