Weak convergence of a sequence of characteristic functions I am trying to produce a sequence of sets $A_n \subseteq [0,1] $ such that their characteristic functions $\chi_{A_n}$ converge weakly in $L^2[0,1]$ to $\frac{1}{2}\chi_{[0,1]}$. 
The sequence of sets 
$$A_n = \bigcup\limits_{k=0}^{2^{n-1} - 1} \left[ \frac{2k}{2^n}, \frac{2k+1}{2^n}  \right]$$
seems like it should work to me, as their characteristic functions look like they will "average out" to $\frac{1}{2} \chi_{[0,1]}$ as needed. However, I'm having trouble completing the actual computation.
Let $g \in L^2[0,1]$, then we'd like to show that 
$$
\lim_{n \to \infty}  \int_{[0,1]} \chi_{A_n} g(x) dx = \int_{[0,1]} \frac{1}{2}\chi_{[0,1]} g(x) dx = \frac{1}{2} \int_{[0,1]} g(x) dx
$$
We have that
$$
\int_{[0,1]} \chi_{A_n} g(x) dx = \sum\limits_{k=0}^{2^{n-1}-1} \int_{\left[ \frac{2k}{2^n}, \frac{2k+1}{2^n}   \right]  } \chi_{A_n} g(x) dx
$$
Now I am stuck, as I don't see how to use a limit argument to show that this goes to the desired limit as $ n \to \infty$. Does anyone have any suggestions on how to proceed? Any help is appreciated! :)

Suggestions: 


*

*First consider the case where $g$ is the characteristic function of an interval.

*Generalize to the case where $g$ is a step function.

*Use density of step functions in $L^2$.

