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In a book, I found the following theorem (slightly simplified here):

Consider $\mathbb{R}$ with the Borel $\sigma$-field.

Let $B$ be the set of all measurable, bounded, real-valued functions on $\mathbb{R}$.

Let $M$ be the set of all finite, signed measures on $\mathbb{R}$.

Let the weak topology on $B$ be defined to be the coarsest topology such that all functions

$B \to \mathbb{R},\quad f \mapsto \int f \mathrm{d} \mu,\quad \mu \in M$,

are continuous.

Theorem: On any $\| \cdot \|_\infty$-bounded subset of $B$, the weak topology and the topology of pointwise convergence coincide.

Proof. Since $M$ contains all one-point probability measures, convergence in the weak topology implies pointwise convergence. Conversely, since sets bounded in $\| \cdot \|_\infty$-norm are also pointwise bounded, we can deduce from Lebesgue's dominated convergence theorem that pointwise convergence implies convergence in the weak topology. $\square$

Now, I do not understand why having the same convergent sequences in this case should imply that the topologies are the same. One direction seems clear: The topology of pointwise convergence is coarser than the weak topology, since the topology of pointwise convergence is the initial topology w.r.t. the family of all functionals $f \mapsto \int f \mathrm{d} \delta_x$.

Yet I have not been able to get rid of the sequences argument in the other direction, and, thinking about how open sets in $B$ look like in the two topologies, I am not even sure anymore that the theorem is correct.

Questions:

(a) Is the theorem correct? (b) Is the proof correct, and, if yes, why?

JohnB
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    A pair of metric spaces with the same convergent series have the same topology. However, the general case is not true. For instance, for the discrete and the cocountable topology in $\mathbb{R}$, a series is convergent iff it is eventually constant. But I am struggling to find a distance to generate the weak topology in your case. – Diego F Medina Sep 13 '17 at 14:43
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    This is not really the weak topology on the Banach space $B$, but another (coarser) weak topology, right? The dual of $B$ is the (larger) space of all bounded signed charges, where a charge is a finitely additive set function (not necessarily $\sigma$-additive). So your weak topology sits somewhere between the $\sigma(B,B')$-topology and the pointwise topology. – Josse van Dobben de Bruyn Dec 08 '20 at 14:40

1 Answers1

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Neither the theorem nor its proof are correct. You can use that the dominated convergence theorem does not apply to nets. To quote t.b.'s answer here:

Here's a cautionary example on the unit interval which you can modify to concoct all sorts of counterexamples:

Let $\Lambda$ be the set of finite subsets of $[0,1]$ ordered by inclusion. For each $\lambda \in \Lambda$ choose a continuous function $f_{\lambda}: [0,1] \to [0,1]$ such that $f_{\lambda}(x) = 1$ for all $x \in \lambda$ and $\int f_{\lambda} \leq \frac{1}{2}$. Then $f_{\lambda} \to 1$ everywhere on $[0,1]$, while $\frac{1}{2} \leq \int (1 - f_{\lambda}) \nrightarrow 0$…

By a slight modification, we get a net $\{f_\lambda\}_{\lambda \in \Lambda}$ in the unit ball of your space $B$ that converges pointwise to $\chi_{[0,1]}$, but with $\int f_\lambda \, d\mu_0 \leq \tfrac{1}{2}$ for all $\lambda \in \Lambda$, where $\mu_0 \in M$ denotes Lebesgue measure on $[0,1]$. (The only modification we need for this, is adding the requirement that $f_\lambda$ is zero outside the interval $[-\tfrac{1}{|\lambda|} , 1 + \tfrac{1}{|\lambda|}]$, in order to guarantee that $\{f_\lambda\}_{\lambda\in\Lambda}$ converges to $0$ outside the interval $[0,1]$.)

To complete the proof, note that the net $\{f_\lambda\}_{\lambda\in\Lambda}$ converges pointwise to $\chi_{[0,1]}$, but not weakly, since $$ \lim_{\lambda \in \Lambda} \int f_\lambda \: d\mu_0 \neq \int \chi_{[0,1]} \: d\mu_0. $$

Josse van Dobben de Bruyn
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