Proof of Riemann integrability 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.) 
 A: 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.
A: First, what are your attempts on the problem? 
A hint:
Recall that there is a theorem which said Riemann integrable and Darboux integrable are equivalent.  Try a proof by contradiction, using the epsilon-partition definition of how it means to be "Darboux integrable".  We should get a contradiction to the definition with a "certain" epsilon.
A: Well the antiderivative of $\frac{1}{x}$ is $\log(x)$ no show what happens when $x\to 0$
$$\int_0^1  f(x)\, \mathrm{d}x= \int_0^\varepsilon f(x) \, \mathrm{d}x + \int_\varepsilon^1 f(x)\, \mathrm{d} x$$
Hence
$$ \int_0^1 f(x)\, \mathrm{d}x= \int_0^\varepsilon f(x)\, \mathrm{d}x+ \ln(1) - \ln(\varepsilon)=\int_0^\varepsilon f(x) \, \mathrm{d}x - \ln(\varepsilon)$$
