- #1
xsw001
- 37
- 0
Suppose f:[a, b]-> R is bounded function
f(x)=0 for each rational number x in [a, b]
Prove Lower Integral <= 0 <= Upper Integral
Proof:
f(x) = 0 when x is rational
both L(f, p) = U(f, P) = 0
and L(f, p) <= Lower Integral <= Upper Integral <= U(f, p)
This function seems like discontinous even though there aren't any information of functional value when x is NOT rational. It looks like that the Intermediate Value Theorem or Mean Value Theorem need to be appplied.
So I have to prove that the Lower Integral <=0, and the Upper Integral >=0. So the function itself has to cross f(x)=0 isn't it?
Any suggestions would be greatly appreciated.
f(x)=0 for each rational number x in [a, b]
Prove Lower Integral <= 0 <= Upper Integral
Proof:
f(x) = 0 when x is rational
both L(f, p) = U(f, P) = 0
and L(f, p) <= Lower Integral <= Upper Integral <= U(f, p)
This function seems like discontinous even though there aren't any information of functional value when x is NOT rational. It looks like that the Intermediate Value Theorem or Mean Value Theorem need to be appplied.
So I have to prove that the Lower Integral <=0, and the Upper Integral >=0. So the function itself has to cross f(x)=0 isn't it?
Any suggestions would be greatly appreciated.
Last edited: