How to prove a property of integrals (Spivak's Calculus - Chapter 13, Problem 16)

Question

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

$\int_{ca}^{cb} f(x)dx=c\int_{a}^{b}f(cx)dx$

Answer 1

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.

Answer 2

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 $c<0$.