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Let $k>0$ and $f: [a, b] \to \mathbb R$ integrable function (on $[a, b]$) Show that the function $g: [{a \over k}, {b \over k}] \to \mathbb R$, $g(x) = f(kx)$, is integrable on the domain $[{a \over k},{b \over k}]$ and

$k \int_{a /k}^{b/k} g = \int_a^bf$

So, am I able to do this with definition of Riemann integrability, or something else, I got kinda messed up doing it

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    $\begingroup$ Use definition of Riemann integral and it should not be difficult. $\endgroup$ – Paramanand Singh Feb 16 '17 at 16:57
  • $\begingroup$ Yeah, it probably should not be difficult. Still can not get it though. $\endgroup$ – repulsive23 Feb 17 '17 at 13:52
  • $\begingroup$ I have provided an answer. $\endgroup$ – Paramanand Singh Feb 17 '17 at 14:30
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I provide an outline of the proof. If $$P=\{x_{0},x_{1},x_{2},\dots,x_{n}\} $$ is a partition of $[a, b] $ then there is a corresponding partition $$P'=\{x_{0}',x_{1}',x_{2}',\dots,x_{n}'\} $$ of $[a/k, b/k] $ where $x_{i} '=x_{i} /k$ and vice versa. Also note that $\Delta x_{i} =k\Delta x_{i}' $ and therefore we have $$S(f, P) =\sum_{i=1}^{n}f(t_{i})\Delta x_{i} = k\sum_{i=1}^{n}g(t_{i}')\Delta x_{i} '= k S(g, P') $$ where $t_{i}' =t_{i} /k$.

Thus for every Riemann sum $S(f, P) $ there is a Riemann sum $S(g, P') $ and vice versa. Since Riemann integral is the limit of Riemann sum, the relation between Riemann sums of $f, g$ is maintained in their integrals also. Thus $$\int_{a} ^{b} f(x) \, dx=k\int_{a/k} ^{b/k} g(x) \, dx$$

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