Show that for $a \ne 0, b \in \Bbb R$ that $\int_{\Bbb R}f(ax+b)d\lambda(x) = \frac1{|a|}\int_{\Bbb R}f(x)d\lambda(x)$ where $\lambda$ is the Lebesgue measure. (Note that both sides are well defined and equal or neither are well-defined).
I have this question for an assignment, but I don't really know where to start. Do I approximate $f$ by simple functions and try to show the assertion for simple functions? I don't need a full answer, but if someone could point me in the right direction or help me get started, that would be great :)
Hint
Suppose $f=\boldsymbol 1_{A}$ where $A$ is measurable. Then, $$\int_{\mathbb R}\boldsymbol 1_A(ax+b)\,\mathrm d \lambda (x)=\int_{\mathbb R}\boldsymbol 1_{\frac{A-b}{a}}(x)\mathrm d \lambda (x)=\lambda \left(\frac{A-b}{a}\right)=\frac{1}{|a|}\lambda (A)=\frac{1}{|a|}\int_{\mathbb R}f(x)\,\mathrm d \lambda (x).$$
Now, try to see what happen when $f$ is simple, i.e. $f(x)=\sum_{i=1}^n\alpha _i\boldsymbol 1_{A_i}$ for $\alpha _i\in\mathbb R$ and $A_i$ measurable. Then use the fact that any positive function is the increasing limit of a sequence of simple function, i.e. there is $(f_n)$ of simple function s.t. $f_n\nearrow f$. And finally, if $f$ is measurable, then $f=f^+-f^-$ where $f^+$ and $f^-$ are positive function (namely, $f^+=f(x)\vee 0$ and $f^-(x)=-(f(x)\wedge 0)$.