Let $\mathcal{I}:=[0,1]$.
Def: A measurable function $\varphi:\mathcal{I}\rightarrow \mathcal{I}$ is said to be uniformly distributed with respect to the Lebesgue measure $\Lambda$ if, for any measurable set $\mathcal{A}\subset \mathcal{I}$, $$ \Lambda_1[\varphi^{\star}(\mathcal{A})]=\Lambda_1(\mathcal{A}) $$ where $\varphi^{\star}(\mathcal{A}):=\{t\in \mathcal{I} \text{ s.t. } \varphi(t)\in \mathcal{A}\cap \mathcal{R}(\varphi)\}$ and $\mathcal{R}(\varphi)$ is the range of $\varphi$.
Consider the measurable functions $\varphi:\mathcal{I}\rightarrow\mathcal{I}$ and $\psi:\mathcal{I}\rightarrow\mathcal{I}$ and assume these are uniformly distributed. For $t\in \mathcal{I}$, construct the function $$ \beta:\mathcal{I}\rightarrow \mathcal{I} \text{ with } \beta(t):=\varphi(\psi(t)) $$
I want to show that $\beta$ is uniformly distributed [from Sagan "Space-Filling Curves" p.110].
Proof (attempt):
We want to show that $\forall \mathcal{A}\subset \mathcal{I}$ with $\mathcal{A}$ measurable $$ \Lambda_1(\beta^{\star}(\mathcal{A}))= \Lambda_1(\mathcal{A}) $$ We have that $$ \begin{aligned} \beta^{\star}(\mathcal{A})&:=\{t \in \mathcal{I} \text{ s.t. } \beta(t)\in \mathcal{A}\cap \mathcal{R}(\beta)\}=\{t \in \mathcal{I} \text{ s.t. } \varphi(t)\in \overbrace{\mathcal{A} \cap \varphi_{\star}(\mathcal{R}(\varphi))}^{\subseteq \mathcal{A}\cap \mathcal{R}(\varphi)} \}\subseteq \{t \in \mathcal{I} \text{ s.t. } \varphi(t) \in \mathcal{A}\cap \mathcal{R}(\varphi)\}\\ &:=\varphi^{\star}(\mathcal{A}) \end{aligned} $$ We know that $$ \Lambda_1(\varphi^{\star}(\mathcal{A}))=\Lambda_1(\mathcal{A}) $$ How can this imply $$ \Lambda_1(\beta^{\star}(\mathcal{A}))=\Lambda_1(\mathcal{A}) $$ ? It can be if and only if $\Lambda_1(\beta^{\star}(\mathcal{A}))=\Lambda_1(\varphi^{\star}(\mathcal{A}))$ ??