Let $f: [0,1] \to R$ be Riemann integrable, and let $a > 0$ be a real number. Show that $ f({x\over a})$ is Riemann integrable on $[0,a]$

Question

My attempt:

Define $g: [0,a] \to R$ as $g(x) = f({x\over a})$. Moreover, let P = {$0,1$} and Q = {$0,a$}. Since $f$ is integrable, it follows that for any $\epsilon >0 \space $ $U(f, P) - L(f,P) < {\epsilon\over a}$. Notice that $\forall x \in [0,a] \exists y\in[0,1] g(x) = f(x)$. Hence, $sup_{x\in [0,a]} g(x) = sup_{x\in [0,1]} f(x)$ and $inf_{x\in [0,a]} g(x) = inf_{x\in [0,1]} f(x)$. So, we have that $U(g,Q) - L(g,Q) = (U(f,P) - L(f,P))a < {\epsilon \over a}a = \epsilon$.

Is this proof correct? Particularly, am I allowed to pick an explicit partition for $f$?

EDIT: If this proof is correct, can someone show a better way to prove this result?

Thanks.

Answer 1

No, your proof is not correct. Given $\varepsilon>0$, you have no reason to assert that, if you take $P=\{0,1\}$, then $U(f,P)-L(f,P)<\frac\varepsilon a$. For instance, take $f(x)=x$, $a=1$ and $\varepsilon=\frac1{10}$.

Answer 2

We just need to prove that $f(x/a)$ is continious $\forall x \in [0,a]$ let $x_0\in (0,a)$ then $lim{x\to x_0}(f(x/a))$ we set $u =x/a$ with $u\to \frac{x_0}{a}=u_0<1$ as $x\to x_0$

so $lim{x\to x_0}(f(x/a))=lim{u\to u_0}(f(u))=f(u_0)=f(\frac{x_0}{a})$ we do tha same for $x\to 0^+ $ and $x\to a^-$ so f(x/a) is continious thus Riemann integrable in $[0,a]$

Answer 3

Before starting a formal proof you have to be aware that the claim is completely obvious: For every partition $P$ of $[0,1]$ which is "good" for $f$ the appropriately scaled partition $P_a$ of $[0,a]$ is "good" for the function $g(y):=f\bigl({y\over a}\bigr)$ $(0\leq y\leq a)$. The rest is "going through the motions".

Answer 4

The simple change that was made slows down (or speed up) $f$ by a constant factor as a function of the NEW $x$, and in turn stretches the "new" domain by the same constant factor. This is just a change of variable. Or a composition of a linear function with the original function $f$.

Use Riemann partitions and sums with the new function, and taking care of the simple substitution, you will get the same sums as for the original $f$.