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Let $D\subseteq \mathbb R^n$ a Lebesgue measurable set, let's call $X$ the space of measurable functions $f:D\to \mathbb R$.

I already know that the almost everywhere pointwise convergence is not metrizable (and also not topologizable).

I want to know if the pointwise convergence on sequences on this set is metrizable, since it is topologizable with the induced topology of $\mathbb R^D$.

I actually proved that any metric $d$ on $X$ that induces the pointwise converges, has a topology $\tau^d$ that contains the restriction of the topology in $\mathbb R^D$, and if they coincide, I probably can conclude that $\tau^d$ is not first countable.

Exodd
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1 Answers1

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The set of all functions in $\mathbb{R}^D$ in the pointwise convergence topoloy is metrisable iff $D$ is countable (for uncountable $D$ indeed first countability fails). A variant of this proof works for the subspace of measurable functions:

Suppose that $U_n$, $n \in \mathbb{N}$, is a local base at the function $f \equiv 0$ (for definiteness, the constant 0 function is certainly in the space).

Then each $U_n$ contains a basic open subset $$B(F_n, r_n) = \{f : D \rightarrow \mathbb{R}: f \text{ measurable }: \forall x \in F: |f(x)| < r \}$$

for some finite subset $F_n \subset D$ and some $r_n > 0$. (This is a standard base for the pointwise convergence topology, relativised to our subspace, corresponding to product open sets that depend on finitely many coordinates (i.e. the set $F_n$). If the $U_n$ form a local base for $f$ so do the sets $B(F_n, r_n)$.

As $D$ is uncountable, $D \setminus \cup_n F_n$ is also uncountable so we have some $p \in D \setminus \cup_n F_n$. Then $f \in B(\{p\}, 1)$ which is open, so should contain some $$B(F_n, r_n) \subset U_n \subset B(\{p\}, 1)$$ for some $n$ But if $g$ is a measurable function that maps $p$ to $2$, and $F_n$ to $0$ (surely these exist, we can even take all other values to be $0$ too, so $g = 2\chi_{\{p\}}$), then $g \notin B(\{p\},1)$, while $g \in U_n$, contradicting the supposed inclusion.

So we cannot have a local countable base for uncountable $D$ in the space of measurable functions in $\mathbb{R}^D$ in the product/pointwise topology. The set of measurable functions is so large we can just mimick the proof for all functions $\mathbb{R}$ to $D$.

Henno Brandsma
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  • There's a hole in this proof: you supposed that $U_n$ contains an open set from the basis of $\mathbb R^D$, but it may be not the case. In fact, to prove that the space is not metrizable, you should take any topology that induces the pointwise convergence. While it's true that it contains the restricted topology of $\mathbb R^D$, it's not easy to see that they coincide. – Exodd Apr 01 '17 at 08:35
  • By the way, I concur that once you proved that the two topologies coincide, it's easy to prove that it is not first countable – Exodd Apr 01 '17 at 08:41
  • If $D$ is uncountable then WLOG $\omega_1+1\subset D.$ So $\mathbb R^D$ with the point-wise convergence topology has a subspace $S={f_x: x\in \omega_1+1} $ where $f_x(y)=1$ if $y\in x $, else $f_x(y)=0. $ Now $S$ is homeomorphic to the $\epsilon$-order topology on $\omega_1+1 ,$ which is not metrizable, because $\omega_1$ is a point with uncountable character. So $\mathbb R^D$ has a non-metrizable subspace, so it is not metrizable. – DanielWainfleet Apr 01 '17 at 09:09
  • @Exodd the topology of pointwise converge is the product topology, they're synomyms. So any $U_n$ contains a basic product set, there is no hole. – Henno Brandsma Apr 01 '17 at 09:32
  • @user254665 It's not simpler to use your proof, it assumes ordinals etc. I wanted to avoid that. You also need to show that the $f_x$ are measurable as well. – Henno Brandsma Apr 01 '17 at 09:36
  • @HennoBrandsma you're assuming that there exists only one topology inducing the pointwise convergence. Maybe it is trivial, but I'm not aware of this result. Could you point a reference? For example http://math.stackexchange.com/questions/76691/example-of-different-topologies-with-same-convergent-sequences shows that it is not true in general – Exodd Apr 01 '17 at 09:36
  • @Exodd nets uniquely determine the topology. Pointwise convergence is defined for nets, normally. If not, pointwise convergence by definition has the (sub)base as I described (more generally $B(f; F, r) = { g : \forall x \in F: |f(x) - g(x)| < r}$ for finite subsets $F$ of the domain (or equivalently singletons if we only want a subbase) and $r>0$. – Henno Brandsma Apr 01 '17 at 09:40
  • @HennoBrandsma. I should have written $S={f_x:x\leq \omega_1+1} $.....WLOG $T= \omega_1+1$ is a measurable subset of $D$. If $x\in \omega_1$ and $U$ is open then$ f_x^{-1}(U)$ is countable or co-countable in $D$. And if $x=\omega_1$ or $\omega_1+1$ then $f^{-1}(U)$ is $T$, or is $T$ minus one point..... I didn't mean to imply this was simpler, It's just familiar to me in this way. – DanielWainfleet Apr 01 '17 at 10:11
  • @user254665 You cannot just assume that $\omega_1 + 1$ is a measurable subset of $D$. $D$ is a real measurable subset so $D$ is metrisable, so cannot contain it topologically. – Henno Brandsma Apr 01 '17 at 10:25
  • @Exodd e.g. see https://www.encyclopediaofmath.org/index.php/Pointwise_convergence,_topology_of or https://en.wikipedia.org/wiki/Pointwise_convergence – Henno Brandsma Apr 01 '17 at 10:26
  • @HennoBrandsma This seems a not so known result. So convergence on nets uniquely determines the topology. In our case, it is true that the ptwise convergence on sequences implies the ptwise convergence on nets? – Exodd Apr 01 '17 at 10:33
  • @Exodd The topology on $\mathbb{R}^D$ is ( I think) not determined by sequences in general. Saying that "sequences converge iff they converge pointwise" is not enough to determine a unique topology on a large product. For nets (genralised sequences) it is enough. And then it is the product topology (restricted to the subspace in question). – Henno Brandsma Apr 01 '17 at 10:50
  • @HennoBrandsma. D does not have to contain $\omega_1$ topologically. It's just an uncountable subset of D. – DanielWainfleet Apr 01 '17 at 22:26