
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. 

Tags 
<, integral, lower, prove, upper 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integral with lower and upper limits  agnieszkam  Calculus  2  March 14th, 2014 05:49 AM 
[SOLVED] Lower bound of an integral  Juliayaho  Real Analysis  0  May 10th, 2013 03:50 PM 
Upper and Lower Sums  valerieangel23  Calculus  1  December 12th, 2012 09:32 AM 
Use upper and lower sum to show that  450081592  Calculus  1  January 13th, 2010 01:15 PM 
Upper and Lower Estimate of Integral  mathlover88  Calculus  1  November 11th, 2009 11:43 AM 