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?