TITLE: Proof of Riemann integrability
QUESTION [5 upvotes]: Let the unbounded function $f(x)$ on the interval $[0,1]$ be defined as
$$f(x)=\left\{\begin{array}{l l}\frac{1}{x}&x\text{ in }(0,1]\\ f(x)=0&x=0\end{array}\right.$$
Show that $f(x)$ is not Riemann integrable.(Hint:infinity is not a real number.)

REPLY [2 votes]: Apparently you want a Riemann sum argument. So take $n\geq 1$ and consider the partition $\mathcal P_n=\{1/n, 2/n,\ldots,(n-1)/n,1\}$ of $[0,1]$. The corresponding Riemann sum of $f$ is
$$
R(f,\mathcal P_n)=\frac{1}{n}\sum_{k=1}^n\frac{1}{\frac{k}{n}}=\sum_{k=1}^n\frac{1}{k}.
$$
Now you recognize the partial sum of the harmonic series. It can be proved without integration that this tends to $+\infty$ by grouping terms as follows
$$
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{8}+\ldots
$$
$$
\geq 1+ \frac{1}{2}+\left( \frac{1}{4}+ \frac{1}{4}\right)+ \left( \frac{1}{8}+  \frac{1}{8}+ \frac{1}{8}+\frac{1}{8}\right)+\ldots=1+\frac{1}{2}+\frac{1}{2}+\ldots
$$
So these Riemann sums tend to $+\infty$ and therefore $f$ is not Riemann-integrable.