Consider the unit circle $\mathbb{S}^1 \subset \mathbb{R}^2$. Let $\nu$ be the uniform measure on $\mathbb{S}^1$. Then there is an orthonormal basis for $L^2(\mathbb{S}^1,\nu)$, namely $\{1,z,\overline{z},z^2,\overline{z^2},\ldots\}$ for which the Riemann-Lebesgue lemma is true. That is, if $g\in L^2(\mathbb{S}^1,\nu)$, then we have $$ \lim_{n \rightarrow \infty} \int_{\mathbb{S}^1} f_n(z)~ g(z)~ d\nu(z) =0, $$ for $f_n(z) = z^n$ or $\overline{z^n}$.
My question: Suppose $\mu$ is an arbitrary Borel probability measure (whose support is not a finite set) on $\mathbb{S}^1$. Is it possible to find an orthonormal basis for $L^2(\mathbb{S}^1,\mu)$, namely $\{f_n\}_{n\geq 1}$, such that for every $g\in L^2(\mathbb{S}^1,\mu)$ $$ \lim_{n \rightarrow \infty} \int_{\mathbb{S}^1} f_n(z)~ g(z)~ d\mu(z)=0? $$
Remark: For $\mu$ supported on a countable set $\{a_n\}_{n\geq 1}$ with $p_n = \mu(a_n) >0$, the natural choice of ONB, namely $\{\frac{1}{\sqrt{p_n}}\delta_{a_n}\}_{n \geq 1}$ works.
Thanks!
Since $\mathscr{B}(\mathbb{S}^1)$ is countable generated, $L_2(\mathbb{S}^1,\mathscr{B}(\mathbb{S}^1),\mu)$ is separable. Once you have a countable dense set $\{\phi_n:n\in\mathbb{N}\}$, you can obtained an orthonormal basis by Gram-Schmidt orthogonalization. In fact, as $\{e^{-\pi n\theta i},n\in\mathbb{Z}\}$ is uniformly dense (by the Stone Weierstrass theorem), it will also be dense in $L_2$. You can then use GS-method to find an orthonormal basis.
Riemann-Lebesgue will follow from Parseval's theorem.