mgfortes wrote:
If f is continuous on (a,b), the f(a) and f(b) can be defined so that f is integrable on [a,b].
My inclination is to say false and to use sin(1/x) as a counterexample. However, since this is a problem I would need to present on the board using "because" as my explanation probably wouldn't be sufficient. How might I explain this accurately and concisely? Any help appreciated!
MF
Everything said above is certainly true...on the other hand, you might be referring to
uniform continuity, since we can say something like this:
A function f is uniformly continuous on the interval (a,b) if and only if it can be defined at the endpoints a and b such that the extended function is continuous on [a,b].
This is sometimes called the "Continuous Extension Theorem". The essential part of the argument to verify it is that if we assume f to be uniformly continuous and the sequence {x_k} is Cauchy, then the image sequence {f(x_k)} is also Cauchy.
Therefore, you may choose to rephrase the original question as follows:
Quote:
If f is continuous on (a,b), then f(a) and f(b) can be defined so that f is integrable on [a,b]
(trivally true, see Outermeasure's post) AND

will be (properly) Riemann integral
(not necessarily true, see Shadow's example).