Show that $f(x) =\frac{1}{x} $ is Riemann integrable Given is a function $f:[1,2] \rightarrow \mathbb{R}$ that is defined as $f(x)=\frac{1}{x}$. I have to show that $f(x)$ is Riemann integrable. 
So what I had was very simple, but I am so unsure about the dissection that I have to pick. 
Let $\epsilon > $ 0 be given en pick $D_n$ = $\left( 1, 1+\frac{1}{n}, 1+\frac{2}{n}, 1+\frac{3}{n},\cdots , 2 \right)$.
Since $f$ is a decreasing function on this interval it's also decreasing on $[t_{k-1}, t_{k}]$, so 
$$\sup_{x\in [t_{k-1}, t_{k}]}f(x) = \frac{1}{t_{k-1}},$$
 and
$$\inf_{x\in [t_{k-1}, t_{k}]} f(x)= \frac{1}{t_k}.$$ 
Now it follows that
$\displaystyle
U(f,D_n ) - L(f,D_n)$ = $\sum_{k=1}^n \left(\frac{1}{t_{k-1}} - \frac{1}{t_k}\right) (t_k - t_{k-1})$ = $\sum_{k=1}^n \left(\frac{1}{t_{k-1}} - \frac{1}{t_k}\right) \frac{1}{n}$. 
After this I started getting serious doubts, it seems that using this dissection won't work on this interval. I am sure it has to do with the interval starting at $1$ instead of the normal $0$. I know that the professor also said something about me having to adjust the dissection for this interval that starts at $1$. But as of now I really have no real idea of how to pick the right dissection. 
So the real question here is how do you pick a good dissection that will works out nicely. Also does it really have to depend that much on the interval starting at $1$? I mean if you slice the interval in equal pieces it shouldn't matter that much right? 
 A: $U(f,D_n)-L(f,D_n)$ telescopes to ${1\over n}\left({1\over 1}-{1\over2}\right)={1\over2n}$. Just pick $n$ big enough to make sure ${1\over2n}<\varepsilon$. If $\pi[1,2]$ is the set of partitions of $[1,2]$ we have 
$$0\le\inf_{P\in\pi[1,2]} U(f,P)-\sup_{P\in\pi[1,2]} L(f,P)\le U(f,D_n)-L(f,D_n)<\varepsilon$$
As $\varepsilon$ is arbitrariliy small we have$\;\inf_{P\in\pi[a,b]} U(f,P)=\sup_{P\in\pi[a,b]} L(f,P)$ which proves $f$ is $R$-integrable.
A: Generally, if $f$ is a monotone (non-decreasing or non-increasing), real-valued function on a closed, bounded interval $[a, b]$, then $f$ is Riemann integrable by a similar argument.
Specifically, fix an arbitrary positive integer $n$, and consider the equal-length partition $P = (x_{k})_{k=0}^{n}$ with
$$
x_{k} = a + k\, \Delta x = a + k \cdot \tfrac{1}{n}(b - a).
$$
If $f$ is non-decreasing, then
\begin{align*}
\sup \{f(x) : x_{k-1} \leq x \leq x_{k}\} &= f(x_{k}), \\
\inf \{f(x) : x_{k-1} \leq x \leq x_{k}\} &= f(x_{k-1}).
\end{align*}
Consequently,
\begin{align*}
U(f, P) - L(f, P)
  &= \Delta x \sum_{k=1}^{n} \bigl[f(x_{k}) - f(x_{k-1})\bigr]
  && \text{$\Delta x = \tfrac{1}{n}(b - a)$ is independent of $k$} \\
  &= \tfrac{1}{n}(b - a)\bigl[f(b) - f(a)\bigr]
  && \text{telescoping sum,}
\end{align*}
which can be made arbitrarily small by taking $n$ sufficiently large.
