Given an orthonormal system of functions {$\phi_n$} in $L^2([0,1])$, we let $E$ be the collection of $x \in [0,1]$ such that the limit $\lim_{n \rightarrow \infty} \phi_n(x)$ exists. Define the function $$\phi(x) = \lim_{n \rightarrow \infty} \chi_E \phi_n(x).$$ Prove that $\phi (x) \equiv 0$.
Fix $x \in [0,1]$. From the way $\phi(x)$ is defined, it seems that we are considering the limit of the sequence $$\chi_E(\phi_1(x)), \chi_E(\phi_2(x)), \chi_E(\phi_3(x)),...,$$ and since $\chi_E(x)$ outputs $1$ when $x \in E$ and $0$ otherwise, we are asking if $\phi_n(x) \in E$ for each $n$. But by the definition of the set $E$, $\phi_n(x) \in E$ implies that $\lim _{k \rightarrow \infty} \phi_k(\phi_n(x))$ exists.
If this interpretation is correct, then we are asking if the sequence $$\phi_1(\phi_n(x)), \phi_2(\phi_n(x)), \phi_3(\phi_n(x)),...,$$ has a limit for a fixed $n$. To prove that $\phi(x) \equiv 0$, we need to show that this sequence DOESN'T have a limit. I don't know how to proceed from here, and how to use the fact that $\phi_n$ is an orthonormal function.
For any measurable set $F$ we have $\int_F \phi =\lim \int_F \chi_E\phi_n$. This is a consequence of uniform integrabilty: The fact that $L^{2}$ norm of $\chi_E\phi_n$ is bounded by $1$ for each $n$ implies uniform integrability. [Ref. Rudin's RCA or any standard text on Probability Theory, e.g. Chung's book]. Now a.e. convergence plus uniform integrability implies convergence in $L^{1}$.
We are almost done. Since $g\equiv \chi_E \chi_F \in L^{2}$ it follows that $\lim \int_F \chi_E\phi_n =\lim \langle \chi_E \chi_F, \phi_n \rangle =0$. So $\int_F \phi=0$. Since $F$ is arbitrary it follows that $\phi =0$ a.e.