Question :

Let $X$ be a topological space and $\mathcal{B}(X)$ its Borel $\sigma$-algebra. Under what conditions is it true that a pointise limit of a sequence of measurable functions $f_n:(E,\mathcal{E})\longrightarrow (X,\mathcal{B}(X))$ remains measurable ?

The following is well-known :

Theorem Let $(E,\cal E)$ be an arbitrary measure space, and let $$f_n:(E,\mathcal E)\longrightarrow(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}}))$$ be a sequence of real-valued measurable functions, where $\overline{\mathbb{R}}=[-\infty,+\infty]$, that converges pointwise to $f:E\to\overline{\mathbb{R}}$. Then $f$ is measurable.

The proof goes as follows : one first proves that for any sequence of real valued measurable functions $g_n$, the maps $\inf_n g_n$ and $\sup_n g_n$ are measurable. What makes the proof work is the observation that for any $c\in\mathbb{R}$, $$ \big[\inf_n g_n < c\big]=\bigcup_{n\in\mathbb{N}}\big[g_n < c\big] $$ (and the analoguous result for $\sup_n g_n$). Thus $\liminf_n g_n$ and $\liminf_n g_n$ must be measurable aswell. In case $f_n$ converges pointwise to $f$, this proves that $f$ is measurable.

The proof hinges on special properties of $\mathbb R$, and cannot readily be transported to more general topological spaces $X$, although the proof likely generalizes to a certain class of order topologies (which ones ?).

Since $\mathcal{B}(\mathbb{R}^d)=\bigotimes^d \mathcal{B}(\mathbb{R})$, it readily follows that for any $d\geq 1$, and any sequence of measurable maps $f_n:E\to\mathbb{R}^d$ that converges pointwise to $f$, the limit function $f$ has to be measurable.

Partial Solution :

What follows is inspired by this answer posted by saz.

Lemma. Let $X$ be a regular topological space, $(\mathcal{O}_i)_{i\in I}$ a basis of its topology, $U$ an open subset of $X$, and $(x_n)$ a sequence in $X$ that converges to $x\in X$. The following are equivalent :

  1. $x\in U$

  2. $\exists~i\in I, \exists~N\in\mathbb{N}$ such that $\overline{\mathcal{O}_i}\subset U$ and $\forall n\geq N,~x_n\in \mathcal{O}_i$

Proof. $(\Longleftarrow)$ If the premise is satisfied, then $x\in\overline{\mathcal{O}_i}\subset U$, thus $x\in U$.

$(\Longrightarrow)$ Suppose $x\in U$ : since $X$ is regular, there exists $V$ open subset such that $$x\in V\subset\overline{V}\subset U$$ Since $(\mathcal{O}_i)_{i\in I}$ is a basis for the topology of $X$, there exists $i\in I$ such that $x\in\mathcal{O}_i\subset V$, and thus $\overline{\mathcal{O}_i}\subset U$. Since $x_n\to x$, there exists $N\in\mathbb{N}$ such that $\forall n\geq N$, $x_n\in\mathcal{O}_i$.

Note. The point of this lemma is that membership of a limit point of a sequence to an open set $U$ is determined by the enventual membership of the $x_n$ to auxiliary open subsets.

Proposition. Let $X$ be a second countable regular topological space. Let $(E,\cal E)$ be a measurable space, and let $f_n$ be a sequence of measurable maps $$(E,\mathcal{E})\longrightarrow (X,\mathcal{B}(X))$$ and let $f:E\to X$ be a map such that $f_n$ converges pointwise to $f$. Then $f$ is measurable.

Proof. Let $(\mathcal{O}_i)_{i\in\mathbb{N}}$ be a countable basis of the topology of $X$. We need only show that for every nonempty open set $U\subset X$, $f^{-1}(U)\in\cal E$. Let $\omega\in E$ : we shall write $x_n=f_n(\omega)$ and $x=f(\omega)$. By hypothesis, $x_n\longrightarrow x$. Therefore $$\begin{array}{rcl} \omega\in f^{-1}(U) & \iff & x\in U \\ & \iff & \exists~ i\in\mathbb{N}, \Big(\overline{\mathcal{O}_i}\subset U ~~\&~~\exists~ N\in\mathbb{N},\forall n\geq N, x_n\in \mathcal{O}_i\Big) \\ & \iff & \displaystyle \omega\in\bigcup_{\substack{i\in\mathbb{N}\\ \overline{\mathcal{O}_i}\subset U}}\bigcup_{N\in\mathbb{N}}\bigcap_{n\geq N}\big[f_n\in\mathcal{O}_i\big] \end{array}$$ Hence, $$f^{-1}(U)=\bigcup_{\substack{i\in\mathbb{N}\\ \overline{\mathcal{O}_i}\subset U}}\bigcup_{N\in\mathbb{N}}\bigcap_{n\geq N}\big[f_n\in\mathcal{O}_i\big]\in\cal E$$ and $f$ is indeed measurable.

  • $\begingroup$ This question is related: math.stackexchange.com/q/1343860/36150 $\endgroup$ – saz Jan 1 '18 at 7:35
  • $\begingroup$ @saz Thank you ! It's truly remarkable how close the two questions are :O So regularity and every closed set has a countable system of open neighborhoods, great! $\endgroup$ – Olivier Bégassat Jan 1 '18 at 14:16

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