Given a partition $P = \{x_0, \ldots, x_{n+1}\} \subset (0,1)$, what is $U(f,P) - L(f,P)$? Simplify that.
Let f:[0,1]-> R be bounded on [0,1] and continuous on (0,1). Prove that f is Riemann integrable on [0,1]. Hint: Show that for any epsilon > 0 there exists a partition P of [0,1] such that U(f,P) - L(f,P) < epsilon.
So f is bounded |f(x)|<=M for some M in R. so the sum(k=0,n)[f(x_k)*(t_k - t_(k-1))] <= sum(k=0,n)[M*(t_k - t_(k-1))
I think it is safe to say that f is uniformly continuous on this interval but not sure I can assume that since its not continuous on the closed interval [0,1] and only continuous on open (0,1). Not sure how to use that it is uniformly continuous (if it even is) to prove that it is Riemann integrable.
I'm at a loss on this. Any help is appreciated. Thanks in advance!
Here is an outline of what one has to do. Suppose $0=p_0<p_1<p_2<p_3=1$
Now you know that the function $f$ is uniformly continuous on $[p_1,p_2]$ so $\displaystyle\int_{{p_1}}^{{p_2}} {f(x)dx}$ does exists.
Now for the trick Pick $p_1$ so that $p_1-p_0 < \dfrac{\varepsilon }{{\left| {M - m} \right| + 3}}$
Do a similar thing for $p_3-p_2$.
Ok Well this is starting to make sense but I am still missing something. So Let P = {p_0, p_1, p_2, p_3} and let P1 = {p_1, p_2} So we know U(f,P1) - L(f,P1) < epsilon. Then U(f,P) - L(f,P) = [M(f,[p_0,p_1]) * (p_1-p_0) + U(f,P1) + M(f,[p_2,p_3](p_3-p_2)] - [ m(f,[p_0,p_1]) * (p_1-p_0) + L(f,P1) + m(f,[p_2,p_3](p_3-p_2) ]. Now let |f(x)| <=S then U(f,P) - L(f,P) <= S(p_1-p_0) + U(f,P1) + S(p_3 - p_2) - [ S(p_1 - p_0) + L(f,P1) + S(p_3-p_2)] = U(f,P1) - L(f,P1) < epsilon. What am I missing here when I look at this sum and how to I use that inequality you provided in choosing p_1 - p-0 and p_3 - p_2. I am missing some thing here when computing U(f,P) - L(f,P)
Thanks again! I really appreciate the help but still confused!