Prove Lower Integral <= 0 <= Upper Integral

[xsw001]

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.

 

[Dick]

There's no IVT if the function isn't continuous. The point is that every interval of nonzero size contains a rational number. So the min of f(x) on the interval MUST be <=0 and the max of f(x) on the interval is >=0. It's a lot less complicated than you think it is.

 

[xsw001]

Oh, I see. Thanks!