Prove that the function f:[0,1]--> R defined by
f(x) = {1, if x=1/n, n is an element of Z+
{0, otherwise
is Riemann integrable on [0,1].
Thanks!
1) Using Lebesgue's Theorem: f is Riemann integrable in [0,1] because it is bounded there and the set of points of discontinuity is of measure zero
2) Directly from the definition: let $\displaystyle \sum\limits_{||\Delta||\rightarrow 0}f(c_1)(x_i-x_{i-1})$ be the Riemann sum of f with some partition $\displaystyle \Delta$, and we take the sum when the number of points of the partitions tend to infinity AND the parameter of the partition, i.e. $\displaystyle ||\Delta||$ goes to zero
Now, the above sum is zero UNLESS the some of the points $\displaystyle c_i$ happen to be $\displaystyle \frac{1}{n}$ for some $\displaystyle n\in \mathbb{N}$, and in that case we'll get $\displaystyle \sum\limits_{||\Delta||\rightarrow 0}f(c_1)(x_i-x_{i-1})=\sum\limits_{\substack{||\Delta||\rightarrow 0\\c_i=\frac{1}{n}}}(x_i-x_{i-1})$
As the last sum is at most a countable sum and as $\displaystyle ||\Delta||\rightarrow 0$, for any $\displaystyle \epsilon>0$ we can choose a partition s.t. $\displaystyle x_1-x_{i-1}<\frac{\epsilon}{2^n}$ , and then $\displaystyle \sum\limits_{\substack{||\Delta||\rightarrow 0\\c_i=\frac{1}{n}}}(x_i-x_{i-1})\leq \sum\limits_{n=1}^\infty\frac{\epsilon}{2^n}=\epsi lon$
Tonio