Let $\phi(x) \in \mathcal{C}^1([0,1])$ be a real valued function such that: $$\begin{cases} \phi'(x) > 0 & \forall x \in [0,1] \\ \phi(0) = 0 \\ \phi(1) = 1. \end{cases}$$ I'm asked to prove that the set $$\left\lbrace{f_n(x) \equiv \sin(n\pi\phi(x)}\right\rbrace_1^{\infty}$$ is complete in $L^2([0,1])$.
Here is my proof. For the sake of conciseness I'll omit some calculations, stating only their final results.
My doubt concerns the legitimacy of the change of variable I perform. Can you tell me if it is right to proceed this way? Also, if there are other mistakes, please point them out.
Let $f$ be a function in $L^2([0,1])$ such that $$ 0 = \left\langle f_n, f \right\rangle_{L^2([0,1])} = \int_0^1 \sin(n\pi\phi(x)) f(x) \,dx \quad \forall n \geq 1.$$ Now, if the change of variable defined by $z(x) = \pi \phi(x)$ (which is possible for the properties of $\phi$) is performed, the former equality becomes: $$ 0 = \int_0^\pi \sin(nz) f(x(z)) \dfrac{dx}{dz} \,dz \quad \forall n \geq 1$$ and, since (from the hypotheses on $f$ and $\phi$) it can be proven that $f(x(z)) \dfrac{dx}{dz} \in L^2([0,\pi]),$ from the completeness of the set ${\left\lbrace{\sin(nz)}\right\rbrace}_1^\infty$ in $L^2([0,\pi])$ and from the hypothesis on $\phi'$ we have: $$f(x(z)) \dfrac{dx}{dz} = 0 \quad \Rightarrow \quad f(z) = 0 \ \text{ a. e. } \; \forall z \in [0, \pi] \qquad \Rightarrow \quad f(x) = 0 \ \text{ a.e. } \; \forall x \in [0, 1]$$ which means that the set ${\left\lbrace{\sin(n\pi\phi(x)}\right\rbrace}_1^\infty$ is complete in $L^2([0,1])$.