Convergence of Indicator function in weak*-topology

Question

Let $\omega_n$ and $\omega$ be measurable subsets of $[0,1]$. Also let indicator function $\chi_{\omega_n}$ converge to $\chi_{\omega}$ in weak*-topology on $L^\infty(0,1)$, that is $$\int_0^1f(x)(\chi_{\omega_n}(x)-\chi_{\omega}(x))dx \to 0, \quad \forall f\in L^1(0,1).$$ What can we extract from this. Are we able to say for instance $$\int_0^1|\chi_{\omega_n}(x)-\chi_{\omega}(x)|^2dx \to 0.$$

Answer 1

As pointed out in the comment above there is no absolute value sign in the definition of weak convergence.

You should also note that $|\chi_{\omega_n}-\chi_{\omega}|^{2}=|\chi_{\omega_n}-\chi_{\omega}|=\chi_{\omega_n \setminus \omega}+\chi_{\omega \setminus \omega_n}$. To see that this tends to $0$ apply the hypothesis with $f=\chi _{\omega}$ and then with $f=\chi _{\omega^{c}}$. I will let you complete the proof. [Will give details if needed].