Question: Let $f:[a,b]\rightarrow R$ be a Riemann integrable function. Let $\alpha>0$ and $\beta \in R$. Then define $g(x)=f(\alpha x + \beta)$ on the interval $I=[\frac {a-\beta}{\alpha},\frac {b-\beta}{\alpha}]$. Show that $g$ is Riemann integrable on $I$.

Attempt: If I sub $I$ into $g(x)$, I get $[f(a), f(b)]$. Therefore, it is equivalent to integrate $f$ on $[a,b]$. Since $f$ is Riemann integrable, $g$ is also Riemann integrable.

Is it okay? I can't find any different way to prove this. If it is wrong, could you give me some other suggestion??

Thank you in advance!

Edit: Let a set of partition, $P=\{a=x_0,x_1...,x_{n-1},x_n=b\}$. Let $m_i=\inf \{f(x):x_{i-1}\le x \le x_i\}$, and $M_i=\sup\{f(x):x_{i-1}\le x \le x_i\}$. Then, $\sum_{i=0}^{n} \Delta x_i m_i=L(P,f)$, (the lower Daboux sum), and $\sum_{i=0}^{n} \Delta x_i M_i=U(P,f)$, (the upper Daboux sum). Then, $\underline{\int_{a}^{b}f} = \sup\{L(P,f): \forall P\}$, and $\overline{\int_{a}^{b}f} = \inf\{U(P,f): \forall P\}$. Since $f$ is Reiemann integrable, $L(P,f)\le \int_{a}^{b}f \le U(P,f) $.

Could you explain how to show $g(x)$ is Riemann integral with the above context?

  • 2
    $\begingroup$ If the argument were correct then it would work for any bijection (or even surjection) $h$ such that $f(h(x))=g(x)$. To produce a proper proof you need to take a Riemann sum for $g$ and show that it is equal to a corresponding Riemann sum for $f$. Since the latter converge so do the former. For this you will need to use that $x\mapsto \alpha x+\beta$ is a homothety and a translation. This means that the shape of the partitions are going to be preserved. $\endgroup$
    – user547557
    Apr 4, 2018 at 3:53

1 Answer 1


Almost, but it is recommended to write out the full details. I will use the Riemann sum criterion, for other criterion should be similar. Denote $L=\displaystyle\int_{a}^{b}f(x)dx$, by guessing, the integral of $g$ should be $L/\alpha$. Given $\epsilon>0$, there is some $\delta>0$ such that \begin{align*} \left|\sum f(c_{i})\Delta x_{i}-L\right|<\epsilon, \end{align*} for all partition $P=\{a=x_{0}<\cdots<x_{n}=b\}$ with $\|P\|<\delta$ and $c_{i}\in[x_{i-1},x_{i}]$. For any partition $Q=\{f(a)=y_{0}<\cdots<y_{n}=f(b)\}$ such that $\|Q\|<\delta/\alpha$, consider the partition $P=\{a=x_{0}<\cdots<x_{n}=b\}$, where $x_{i}=\alpha y_{i}+\beta$, then $\|P\|<\delta$, we get \begin{align*} \left|\sum g(d_{i})\Delta y_{i}-\dfrac{L}{\alpha}\right|&=\left|\sum f\left(\alpha d_{i}+\beta\right)\dfrac{\Delta x_{i}}{\alpha}-\dfrac{L}{\alpha}\right|\\ &=\alpha^{-1}\left|\sum f(\alpha d_{i}+\beta)\Delta x_{i}-L\right|\\ &<\alpha^{-1}\epsilon. \end{align*}

  • $\begingroup$ What does this notation $||.||$ mean? $\endgroup$ Apr 4, 2018 at 4:08
  • $\begingroup$ The mesh of a partition, that is, $\|\cdot\|=\max_{1\leq i\leq n}\{|x_{i}-x_{i-1}|\}$. $\endgroup$
    – user284331
    Apr 4, 2018 at 4:09
  • $\begingroup$ The definition of Riemann integrable here is by Riemann sum version. If you use Cauchy criterion, the same idea still goes through because of the preserving map $\alpha(\cdot)+\beta$. $\endgroup$
    – user284331
    Apr 4, 2018 at 5:35
  • $\begingroup$ Do you mind if I ask you to explain with the context, which I just edited? I am not familiar with some concepts in your explanation, so I don't clearly understand it. $\endgroup$ Apr 4, 2018 at 6:15
  • $\begingroup$ So you are using Cauchy criterion that $U(P,f)-L(P,f)<\epsilon$ implies integrability. $\endgroup$
    – user284331
    Apr 4, 2018 at 6:18

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