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 October 29th, 2010, 07:05 PM #1 Newbie   Joined: Oct 2010 Posts: 25 Thanks: 0 Prove Lower Integral <= 0 <= Upper Integral 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 needs 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.
 October 29th, 2010, 07:27 PM #2 Newbie   Joined: Oct 2010 Posts: 25 Thanks: 0 Re: Prove Lower Integral <= 0 <= Upper Integral Never mind, I got it. 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.

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