How to prove a property of integrals (Spivak's Calculus - Chapter 13, Problem 16) I'm trying to self study a bit of rigorous calculus before starting university and I would love some help with this problem.
Prove that 
$\int_{ca}^{cb} f(x)dx=c\int_{a}^{b}f(cx)dx$
Thank you! 

As someone said, by the situation of this problem on the book, it should be proven with Riemann Sums, thanks for the help anyway!

As requested in the comments, I post my approach.
We define the partition of [a,b] as P={$t_0,t_1, ...,t_n$}
We know that when n goes to infinity and the limit of $t_i-t_{i-1}$ goes to $0$ the Lower and Upper sums converge to the integral of the function (assuming necessary conditions)
Then, we take $L(f,P)=\sum_{i=1}^{n}f(t_{i-1})(t_i-t_{i-1})$ as $\int_a^bf(x)$
Now we define $L(f,P)'=\sum_{i=1}^{n}f(c*t_{i-1})(t_i-t_{i-1})$ that would be equal 
to $\int_a^bf(cx)$
and $P'=${$ct_o,ct_1,....,ct_n$} the partition of the interval [ca,cb]
As done before, we take 
 $L(f,P')''=\sum_{i=1}^{n}f(c*t_{i-1})(c*t_i-c*t_{i-1})=\int_{ca}^{cb}f(x)dx=$$=c*\sum_{i=1}^{n}f(c*t_{i-1})(t_i-t_{i-1})$
$=c*L'=c*\int_{a}^{b}f(cx)dx$
I hope you can check that and correct me if im wrong.
Thanks for your attention and help. 

Here it comes what I expect to be the last try.
Lets define all the things we are going to use in the proof.
The partitions we are going to use are (as before). 
$P=${$t_0,t_1, ...,t_n$} (partition of [a,b]) and $P'=$ {$ct_o,ct_1,....,ct_n$} (partition of [ca,cb])
As recommended in the comments, lets define $g(x)=cf(cx)$
Lets call $M_i$ the maximum of $g(x)$ in the interval [$t_{i-1},t_i$] and $m_i$ the minimum in the same interval.
$L(g,P)=\sum_{i=1}^n m_{i-1} (t_i-t_{i-1})<\int_a^bg(x)dx$
$L(f,P')=\sum_{i=1}^n m_{i-1}'(ct_i-ct_{i-1})<\int_{ca}^{cb}f(x)dx$
As we are integrating in the interval [ca,cb] and $f(cx)=g(x)/c$ the minimums of the function $f$ can be expressed as $m_{i-1}/c$ then: 
$L(f,P')=\sum_{i=1}^n m_{i-1}/c*(ct_i-ct_{i-1})=\sum_{i=1}^n m_{i-1}(t_i-t_{i-1})=L(g,P)<\int_a^bg(x)$ 
Doing the same substitution with the maximums we get from: 
$U(g,P)=\sum_{i-1}^n M_{i-1}(t_i-t_{i-1})>\int_a^b(g(x)dx$ 
$U(f,P')=\sum_{i-1}^n M_{i-1}'(ct_i-ct_{i-1})=\sum_{i-1}^n M_{i-1}(t_i-t_{i-1})=U(g,P)$ 
By definition: $L(g,P)<\int_a^bf(x)dx<U(g,P)$ and only $\int_a^bf(x)dx$ satisfy this inequality 
So after showing that the lower and uppers sums are equal, we conclude that 
$\int_{ca}^{cb}f(x)dx=\int_{a}^{b}g(x)dx$ 
 So we get
$\int_{ca}^{cb} f(x)dx=c\int_{a}^{b}f(cx)dx$
 A: Big Hint, without giving away the whole solution: 
You're on the right track, mostly, but you've got some notational problems. I suggest that you define $g(x) = c \cdot f(cx)$, just to give it a name. And for any partition $P = \{t_0, \ldots, t_k\}$ of $[a, b]$, you consider the partition $P' = \{ ct_0, \ldots, ct_k \}$ of $[ca, cb]$. 
What I mean here is that you consider "prime" as an operation that takes a partition of $[a. b]$ and produces a partition of $[ca, cb]$. Clearly there's another operation... call it $\circ$, that does the opposite: given a partition 
$Q = \{t_0, \ldots, t_k\}$ of $[ca, cb]$, there's a corresponding partition 
$Q^\circ = \{\frac{1}{a} t_0, \ldots, \frac{1}{a} t_k \}$. And if you consider $(P')^\circ$, you get back $P$, for any partition $P$ of $[a, b]$.
Now your goal is to show that two integrals are equal. Here's how to do that: 


*

*Show that every upper sum for one of them is also an upper sum for the other, and vice versa. Note that each upper sum is just a number, so we're trying to show that two sets of real numbers are the same. 

*show the same thing for lower sums. 
Once you've done that, you know the integrals are equal, for they are the least-upper-bound of the two identical sets (and the greatest lower bounds of two other identical sets). 
Whew. Now with all that notation and stuff in hand, consider, for some partition $P$ of $[a, b]$, the lower sum
$$
s = L(P, g).
$$
Can you think of some partition of $[ca, cb]$ with the property that the lower sum for $f$ with respect to that partition is also $s$? If so, write it down and prove that they're equal (which you've almost done, although a bit garbled, in your question). 
Now take that idea and run with it to complete the two bulleted tasks. 
A: The claim holds trivially when $c=0$ as each side is equal to $0$. I will show the following for $c>0$.

If $g(x) =f(cx) $ is Riemann integrable on $[a, b] $ then $f(x) $ is Riemann integrable on $[ca, cb] $ and the result in question holds.

As in question, let $P$ denote a partition of $[a, b] $ and $P'$ be the corresponding partition of $[ca, cb] $. Since $g$ is Riemann integrable on $[a, b] $, for every $\epsilon >0$ there is a $\delta'>0$ such that for any partition $P$ of $[a, b] $ with norm $||P||<\delta'$ the Riemann sum $$S(P, g)=\sum_{i=1}^{n}g(\xi_{i})(t_{i}-t_{i-1}) ,\,\xi_{i}\in[t_{i-1},t_{i}]$$ satisfies $$\left|S(P, g) - \int_{a} ^{b} g(x) \, dx\right|<\epsilon/c\tag{1}$$ Now we can see that a Riemann sum for $f $ on $[ca, cb] $ is $$S(P', f) =\sum_{i=1}^{n}f(\xi_{i}')(t_{i}'-t_{i-1}')=c\sum_{i=1}^{n}g(\xi_{i})(t_{i}-t_{i-1})=cS(P, g) ,\,\xi_{i}'=c\xi_{i}$$ so that each Riemann sum for $f$ is equal to $c$ times a Riemann sum for $g$ and vice versa. Multiplying $(1)$ by $c$ we can see that $$\left|S(P', f) - c\int_{a} ^{b} g(x) \, dx\right|<\epsilon $$ for all partitions $P'$ of $[ca, cb] $ with norm $||P'||<c\delta'=\delta$. It follows that $$\int_{ca} ^{cb} f(x) \, dx=c\int_{a} ^{b} f(cx) \, dx$$ The converse of the above result is proved in similar manner and the proofs can be easily adapted for the case when $c<0$.
