# Real Analysis Riemann Integral Proof I'm so lost on what to do, I know that I can show that f is Riemann integrable on [0,1] by showing it is Darboux integrable, but am stuck on showing U(f)=L(f) so that this is true. I think I have to use the information given by the sequence but am not sure how to go about this.

• It may be useful to note that convergent sequences of real numbers are nowhere dense. – user328442 Dec 7 '18 at 1:53

## 1 Answer

I think your best bet for showing that this function is Riemann integrable will hinge on two points of considereation:

1) Since the sequence $$(x_n)$$ converges, there is only one limit point of the set of the values $$\{x_n\}$$, say $$L$$.

2) The values of Upper Darboux sums monotonically decrease with parition refinements, and the values of Lower Darboux sums monotonically increase with partition refinements.

Computing the Lower Darboux integral is easy; you just get back $$0$$ since the infimum of $$f$$ on any non-degenerate interval containing an $$x_n$$ must necessarily contain a point that is not an $$x_n$$.

Pick an arbitrary partition of $$[0,1]$$, and then refine it to include as endpoints $$x_n-\delta$$ and $$x_n+\delta$$ for some small $$\delta$$ and some initial segment of the sequence of $$(x_n)$$, possibly also taking into account the location of the limit point $$L$$. Use that refinement to argue that the Upper Darboux integral must be less than $$\epsilon$$ for any positive $$\epsilon$$.