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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 at 1:53

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 $\epsilon$.

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