Suppose that $F$ is a distribution function. Denote by $\mu_F$ the measure on $\mathbb{R}$ induced by $F$.
Suppose that $a>0$. Define a new distribution function $F_a$ by $F_a(x):= F(ax)$, and denote the corresponding measure by $\mu_a$.
Let $u \in \mathscr{L}^1(\mu_F)$, with $u(a) > 0$.
My question: Is the following true. If so, how do you prove it? If not, what is the correct change of variables result?
$$ \int_{\mathbb{R}} u(x) \mu_a(dx) = \frac{1}{u(a)}\int_{\mathbb{R}} u(x) \mu_F(dx).$$
Or in more suggestive notation, is this true:
$$ \int_{\mathbb{R}}u(x) F(a\cdot dx) =\frac{1}{u(a)}\int_{\mathbb{R}} u(x)F(dx).$$
Many thanks for your help.
None of the above. If the PDF of $X$ is $F$ then the PDF of $X_a=a^{-1}X$ is $F_a$ (can you check this?) hence $$ \int u\mathrm d\mu_a=E(u(X_a))=E(u(a^{-1}X))=\int u(a^{-1}x)\mathrm d\mu_F(x). $$