Suppose that $f$ is Lebesgue integrable on the interval $[0,a]$ and $g$ is a function such that $g(x)=f(kx)$, then is $g$ integrable on the interval $[0,\frac{a}{k}]$.
I know that since $f$ is integrable on $E=[0,a]$, then $\int_{E}f<\infty$ and $\int_{E}f=\sup_{\varphi \leq f}\int_{E}\varphi$, where $\varphi$ is a simple function. Since $\varphi$ is simple, then I can write $\varphi=\sum\limits_{k=1}^{n}a_k\chi_{E_k}$.
Where would I go from here to show that $g$ is Lebesgue integrable? I'm totally stuck.
Best Answer
$$ \forall x \in [0, a] \quad \phi(x) \le f(x).$$ Hence $$ \forall t \in [0, a/k] \quad \phi(kt) =\sum_{i=1}^n a_i \chi_{E_i}(kt) = \sum_{i=1}^n a_i \chi_{E_i/k}(t) = \frac{1}{k}\sum_{i=1}^n a_i\chi_{E_i}(t) \le g(t).$$
Then $$ \frac{1}{k}\sum_{i=1}^n a_i\chi_{E_i}(t) \le g(t).$$ implies $$\frac{1}{k}\int f dx =\int g dx .$$