# weak closedness of a set with bounded functions

Let $$A$$ be a compact topological space equipped with the Borel $$\sigma$$-algebra, and $$X=B_b(A)$$ be the vector space of bounded measurable functions. Let $$Y=\mathcal M(A)$$ be the vector space of finite signed measure on $$A$$. Define the dual pair $$<\cdot, \cdot>$$ between $$(X,Y)$$ such that $$=\int_A f(a)\mu(da)$$. Let $$\sigma(X,Y)$$ be the weakest topology such that for all $$\mu\in Y$$, the linear map $$X\ni f\mapsto \in \mathbb R$$ is continuous.

Define the set $$U=\{f\in X\mid \sup_{a\in A}|f(a)|\ge 1\}$$. Is the set $$U$$ closed in the $$\sigma(X,Y)$$ topology? I am not sure how to proceed to prove or disprove the claim.

No, assuming $$A$$ is $$T_1$$ and infinite. Indeed, just choose a sequence of distinct points $$\{a_n\}_{n \in \mathbb{N}} \subset A$$. Then $$f_n = 1_{\{a_n\}} \in U$$, but $$f_n \to 0$$ pointwise, so as they are uniformly bounded by $$1$$, by bounded convergence theorem, $$\langle f_n, \mu \rangle \to 0$$ for all $$\mu \in Y$$, i.e., $$f_n \to 0$$ in $$\sigma(X, Y)$$ topology. But $$0 \notin U$$.
• Is $f_n \rightarrow 0$ pointwise is enough ? Shouldnt the convergence be in something like sup norm. Although for the given set of functions it seems to work. Mar 31 at 1:47