So I have a sequence of sets $A_n = \bigcup_{k=0}^{2^{n-1}-1}[\frac{2k}{2^n},\frac{2k+1}{2^n}]$, let $f_n$ be the characteristic function of $A_n$, it is intuitively clear that $f_n$ converge to $1/2$ weakly on $L^2([0,1])$ with the Lebesgue meausre $\mu$, to show this, I think it is enough to show, for any $E \subset [0,1]$ measurable, that $$ \sum_{k=0}^{2^{n-1}-1}\mu(E \cap [\frac{2k}{2^n},\frac{2k+1}{2^n}]) \to \frac{1}{2}\mu(E) $$ I am not sure how to proceed from here, any suggestions is appreciated.
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1If you are still interested in this question, you should have a look at http://math.stackexchange.com/questions/283737/weak-convergence-of-a-sequence-of-characteristic-functions. – gerw Feb 13 '16 at 14:32