Question related to Riemann sums, sups, and infs of bounded functions

[AxiomOfChoice]

Can someone give me an example of a bounded function f defined on a closed interval [a,b] such that f does not attain its sup (or inf) on this interval? Obviously, f cannot be continuous, but for whatever reason (stupidity? lack of imagination?) I can't think of an example of a discontinuous, bounded function for which \sup\limits_{x\in [a,b]} f(x) \neq \max\limits_{x\in [a,b]} f(x).

 

[lavinia]

Can someone give me an example of a bounded function f defined on a closed interval [a,b] such that f does not attain its sup (or inf) on this interval? Obviously, f cannot be continuous, but for whatever reason (stupidity? lack of imagination?) I can't think of an example of a discontinuous, bounded function for which \sup\limits_{x\in [a,b]} f(x) \neq \max\limits_{x\in [a,b]} f(x).

f(x) = x except at the end points where it equals (a+b)/2

 

[AxiomOfChoice]

Hmmm...I think I may have just thought of one! On [0,1], define f(1) = 0, and let
<br /> f(x) = \sum_{n = 0}^\infty \frac{n}{n+1} \chi_{\left[ \left. \frac{n}{n+1},\frac{n+1}{n+2} \right) \right.}(x).<br />
Then f(x) \leq 1 for all x, \sup\limits_{x\in[0,1]} f(x) = 1, but f(x) \neq 1 for all x\in [0,1]. And, if I'm not mistaken, f is still Riemann integrable, since there are only countably many points at which it is discontinuous. Can someone confirm if this is actually true?

 

[AxiomOfChoice]

f(x) = x except at the end points where it equals (a+b)/2

This is much easier than my example. Thanks!