Compactness of set of indicator functions Let $\chi_A(x)$ denote an indicator function on $A\subset [0,1]$. Consider the set
$$K=\{\chi_A(x): \text{ A is Lebesgue measurable in }[0,1]\}.$$
Is this set compact in $L^\infty(0,1)$ with respect to weak-* topology? How about weak topology on $L^\infty(0,1)$? 

Using Banach-Alaoglu, the set $K$ is bounded and closed in $L^\infty(0,1)$ so it is compact with respect to weak-* topology. It is closed, because if $\chi_{A_1}$ converges to $f$ in $L^\infty(0,1)$, then $f$ has to be an indicator function.
 A: This set is not closed in the weak* topology. Indeed, for each $n$ consider the set $A_n=: [0, \frac{1}{2n})\cup [\frac{2}{2n}, \frac{3}{2n}] \cup\dots\cup [\frac{2n-2}{2n}, \frac{2n-1}{2n})$, i.e. the sum of intervals that covers half of the interval. I claim that the sequence $\chi_{A_n}$ converges to the function $\frac{1}{2}$ in the weak* topology. Indeed, since the sequence is uniformly bounded, it suffices to check the convergence on a dense subset of $L^1$; we will take the continuous functions. Continuous functions on $[0,1]$ are uniformly continuous, so if we fix a function $f$, then for sufficiently big $n$ it will be almost constant on every interval of the form $[\frac{k}{n}, \frac{k+1}{n})$, so the integral on $[\frac{2k}{2n}, \frac{2k+1}{2n})$ will be almost equal to the integral on $[\frac{2k+1}{2n}, \frac{2k+2}{2n})$, up to $\frac{\varepsilon}{n}$, where $\frac{1}{n}$ is coming from the length of the interval. It follows that the integral on $A_n$ is, up to $\varepsilon$, equal to $\frac{1}{2} \int_{0}^{1} f(x) dx$.
On the other hand, this sequence does not have a convergent subnet in the weak topology. Indeed, if it did, it would have to converge to $\frac{1}{2}$. Recall that the weak closure of a convex set is equal to its norm closure. It means that if I take the convex hull of the functions $\chi_{A_n}$, then I can find a sequence of elements converging uniformly to $\frac{1}{2}$. Note that $\chi_{A_n}$ vanishes on the interval $[\frac{2n-1}{2n}, 1]$, so any element of the convex hull vanishes on an interval containing $1$. It follows that the distance of any element in the convex hull to $\frac{1}{2}$ is at least $\frac{1}{2}$, so there can be no convergence.
