Main Idea
Because my construction is somehow not friendly to follow, I'll give a broad summarization of my idea.
So, in our main question, we have to deal with a sequence of functions. Not only that, the sequence convergence is also involved. Given the complexity of directly dealing with that object. I manoeuvre a new sequence permitting us to translate all of our necessary information into it, hence also our question. Then after translation, our analysis of the sequence of functions is translated into an analysis of a specific function.
So you can see in next section, under my transformation, the continuous point of the new function $F$ is "roughly" the point of interest in our main question. And as we know, the set of continuous point of a function is measurable, hence we have the conclusion.
The presence of "Roughly" in previous paragraph is due to the fact that those continuity point of $f$ is also the continuity point of $f$ ( I have proved the vice versa in main answer).
So, considering the set of countinuity of $F$ is entirely true. However, some modifications can be made in the proof of the measurability ( for the set of continuity) in order that we can have the desired property.
The organization of my answer is follows:
- The construction of new space
- The construction of new function $F$ that contains all information of $(f_n)$
- The equivalence of the set of continuity of $F$ and our set of interest ( intersects with the set of continuity of $f$)
- The modication needed for the measurability of new set
$$\tilde{Z}:= \{ p=(\infty, x) : \forall p_n \in P \text{ such that}: p_n \rightarrow p \text{ and } p_W(p_n)<+\infty \text{ we have } F(p_n) \rightarrow F( p) \}$$
Main Answer
Here is my assumption on $X$:
For the convenience of showing the main idea of my proof, let's change $Y$ for now to $Y'=Y\cap V$, where $V$ is the set of continuity of $f$
Measurability of $Y'$
Let :
- $W:=\overline{\mathbb{N}^*} =\mathbb{N^*}\cup \{ \infty\} $, the extended set of positive natural numbers.
- $d_W$ denotes a metric on $W$ which is defined as $d_W(m,n)=\left| \frac{1}{m}-\frac{1}{n}\right|$
- For consitency, $d_X$ denotes the metric of $X$ instead of $d'$
- $f_{\infty} := f$
- $P:=W\times X$ denote the product metric space with the product metric $d_P:= d_W+d_X$
Remark 1: The metric $d_P$ is equivalent to the usual product metric. I use that one instead of the usual one just because of its convenience for the thought process.
Remark 2 $W$ is a separable space ( though it is not separated).
Remark 3 Because $W$ and $X$ are seperable, we have $\mathcal{B}(P)= \mathcal{B}(W)\times \mathcal{B}(X)$
The following is the main argument of our proof.
Let $F$ be a function on $P$ taking value in $E$, defined as follows:
$$ \begin{align} F:& W\times X & \longrightarrow & E \\ &(n,x) & \longmapsto & f_n(x)\end{align} $$
For all measurable set $B \in \mathcal{B}(\mathbb{R})$, we have:
$$ F^{-1}(B)= \bigcup_{ n \in \overline{N^*}} \left( \{n\}\times f^{-1}_n(B) \right)$$
So $F$ is measurable, thus, its set of continuity, denoted by $Z$ for now, is measurable.
Let $U \subset X$ be the set that satisfies : $ \{\infty\} \times U = Z\cap \left( \{\infty\}\times X \right)$
Clearly, the $RHS$ of equality above is measurable. And as the Borel which is defined on $P$ is a product Borel, thus its projection $p_X: P \rightarrow X$ is measurable, which implies the measurability of $U$.
Besides, the definition of $U$ can be requoted as follows:
$$U= \left\{ x \vert \forall ( n_k, x_{k}) \xrightarrow{ d_P} (\infty, x) \text{ we have } f_{n_k}(x_k) \rightarrow f(x) \right\} $$
I'll prove that $U=Y'$
As the condition for a point to be in $U$ is stronger than the one for $Y'$, we have the inclusion $ U \subset Y'$.
On the other hand, if $x$ is a point in $Y'$ and $(p_k)=( n_k, x_{k})$ is an abitrary sequence of points in $P$ such that $ p_k \longrightarrow (\infty, x)$ .
We can either choose a subsequence $(p_{k_m})$ that converges to the same limit and that:
- either, $n_{k_1}<n_{k_2}<n_{k_3}<...$.
- or $+\infty= n_{k_1}=n_{k_2}=n_{k_3}=...$
Remark Such a choice is possible due to the way we define $d_W$
In the first case, we fill that subsequence so have a new sequence $(q_i)=(i,y_i)$ defined as follows:
- $y_i= x_{k_m} $ if $n_{k_m} \le i < n_{k_{m+1}}$
- $y_i$ be any point if $i<n_{k_1}$
$\Rightarrow y_i \rightarrow x$ (due to the convergence of $(p_{k_m})$)
Due to the property of $Y'$, we imply that :
$$\begin{align} & F(q_i) &\longrightarrow & F(( \infty, x))\\ \Rightarrow & F(p_{k_m}) &\longrightarrow & f(x) \end{align} $$
The second case, the limit is achieved by definition of $V$.
So, for any sequence $( F(p_k) )$ , we can find a subsequence that converges $f(x)$.
Thus $F(p_k) \longrightarrow f(x)$ (In fact, we have to take a subsequence of an arbitrary subsequence but it doesn't matter because of our freedom in choosing $(p_k)$ ) .
$\rightarrow x \in U$
$\rightarrow Y' \subset U$
Thus $Y'=U$.
Conclusion : $Y'$ is measurable
Discussion :
- We see that we can even prove that $Y'$ is $G_\delta$
- The assumption is rather weak because most of the spaces with which we usually deal are separable. (The separability permit us to construct a lot of structures on topological space)
- The assumption is essential for our proof because we need a link between the product Borel and the Borel of product metric.
Measurability of Y
As shown above, the continuity of $F$ at a point $p=(\infty,x) \in P$ also contains the continuity of $f$ at point $x$. This hinders us from having our desired conclusion. However, with some modifications in the proof for the measurability of $Z$, we can relax that constraint.
Let :
$$ U_n = \{ p \in P : \exists \delta_p >0 s.t \forall y ,z \in B(p,\delta_p) \text{ and } p_W(y)< +\infty, p_W(z)<+\infty \text{ we have: } \\ d_X(F(y),F(z))<1/n \}$$
where $p_W: P \rightarrow W$ is the projection on $W$.
We can see that this set is open because for any $p \in U_n$ and for any $w \in B(p, \delta_p)$, $w$ is in $U_n$ with $\delta_{w}:= \delta_p- d_P(p,w)>0$.
Then, let:
- $\tilde{Z}:= \bigcap_{n \ge 1} U_n$
- $\tilde{U} := p_X\left( \tilde{Z}\cap (\{\infty\}\times X ) \right)$
By the same procedure as above, we can prove that $\tilde{U}=Y$.
