Proving dilation property of Riemann integral 
Prove: If $ f $ is Riemann integrable on the interval $[a,b]$ then for $c>0$,
  $$\int^b_a f(x)\ \mathrm dx = \frac1c \int^{cb}_{ac} f(\frac xc)\ \mathrm dx.$$

I can't come up with a proof. What's the best way to think about this?
 A: Let ${\cal P}$ be the collection of partitions of $[a,b]$ and let
${\cal P'}$ be the collection of partitions of $[ca,cb]$.
Let $P=(a=x_0,x_1,...,x_n=b) \in {\cal P}$. Then
$P'=(cx_0,...,cx_n) \in {\cal P'}$.
Let $\phi(x) = f({x \over c})$.
So, you have
$U(f,P) = \sum_k (\sup_{t \in [x_k,x_{k+1}]}f(t) ) (x_{k+1}-x_k)$
\begin{eqnarray}
U(\phi,P') &=& \sum_k (\sup_{t \in [cx_k,cx_{k+1}]}\phi(t) ) (cx_{k+1}-cx_k) \\
&=& c\sum_k (\sup_{t \in [x_k,x_{k+1}]}\phi(ct) ) (x_{k+1}-x_k) \\
&=& c\sum_k (\sup_{t \in [x_k,x_{k+1}]}f(t) ) (x_{k+1}-x_k) \\
&=& cU(f,P)
\end{eqnarray}
Exactly the same sort of argument shows that $L(\phi,P) = c L(f,P')$.
It should also be clear that we can do the reverse, if we have a partition $P' \in {\cal P'}$, then we can make a partition $P \in {\cal P}$ with the same properties.
It follows that $ \inf_{\pi' \in {\cal P}'} U(\phi, \pi') = c\inf_{\pi \in {\cal P}} U(f,\pi)$, and similarly,
$ \sup_{\pi' \in {\cal P}'} L(\phi, \pi') = c\sup_{\pi \in {\cal P}} L(f,\pi)$.
By definition, we have $\sup_\pi L(f,\pi) = \int_a^b f = \inf_\pi U(f,\pi)$, and similarly for $\int_{ca}^{cb} \phi$, and the result follows from this.
