# 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.

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.