I'm trying to solve the following problem: Given the Lebesgue measure in $\mathbb{R}$:
$\forall a>0, \forall b\in\mathbb{R}$ and for all $A$ Lebesgue measurable set, show that: $\lambda(aA+b) = a\lambda(A)$.
Conclude that $f\in\mathcal{L}^1$ iff $g(x) = f(ax+b)\in\mathcal{L}^1$, and that:
$$\int f(ax+b) d\mu(x) = \frac{1}{a}\int f(x)d\mu(x)$$
I have no idea where to start with any of the parts of the problem, and would appreciate any help or insight.
Notice that $\lambda$ is translation invariant essentially by definition. This can be shown by translating the open covers of a given set $A$ to $x + A$ for any $x \in \Bbb R$. Likewise you can do this for scaling.
Once you show this, establish the equality in question for indicator functions, where it follows immediately from the first part. Then take use the linearity and the fact the simple functions are dense in $L^1$.