If a sequence converges in $L^2$ and we compose every function with a non-singular function, does it still converge? Let $F\colon (0,1)\rightarrow (0,1)$ be a non-singular function with respect to the Lebesgue measure $\mu$ (so $\mu\sim\mu \circ F$). Let $\{ f_n : n \in \mathbb{N} \} \subset L^{2}([0,1])$ be a sequence of simple integrable functions and $f\in L^{2}([0,1])$ such that $f_n\to f $ in the $2$-norm. Is it correct that also $f_n\circ F\to f\circ F$ ?
If not, what are the conditions on $F$ such that this implication is correct?
 A: Here are some counterexamples. Consider the functions 
$$
f_n(x)=x^{-a(n)}-x^{-b(n)},\qquad F(x)=x^2.
$$
Assume that $0<a(n)<\frac14$, $a(n)\to\frac14$, $0<b(n)<\frac14$ and $b(n)\to\frac14$. Then,
$$
\|f_n\|_2^2=\frac1{1-2a(n)}+\frac1{1-2b(n)}-\frac2{1-a(n)-b(n)}\to0,
$$
hence $f_n\to f$ in $L^2$ with $f=0$. Furthermore, $f\circ F=0$ and $f_n\circ F(x)=x^{-2a(n)}-x^{-2b(n)}$, hence
$$
\|f_n\circ F\|_2^2=\frac1{1-4a(n)}+\frac1{1-4b(n)}-\frac2{1-2a(n)-2b(n)}.
$$
Let $A(n)=1-4a(n)$ and $B(n)=1-4b(n)$, then $\|f_n\circ F\|_2^2=C(n)$ with
$$
C(n)=\frac1{A(n)}+\frac1{B(n)}-\frac4{A(n)+B(n)}=\frac{(A(n)-B(n))^2}{A(n)B(n)(A(n)+B(n))}.
$$
If $nA(n)\to\alpha$ with $\alpha>0$ and $nB(n)\to\beta$ with $\beta>0$, $\|f_n\circ F\|_2^2=\gamma n+o(n)$ with $\gamma\ne0$ as soon as $\alpha\ne\beta$, in which case $f_n\circ F$ does not converge in $L^2$ to $f\circ F=0$ since $\|f_n\circ F\|_2^2\to\infty$.
Likewise, if $n^2A(n)=1+o(1/n)$ and $n^2B(n)=1+c/n+o(1/n)$, $\|f_n\circ F\|_2^2\to c^2/2$ hence $f_n\circ F$ does not converge in $L^2$ to $f\circ F=0$ although the sequence $(f_n\circ F)$ converges pointwise to $0$ and is bounded in $L^2$.
