TITLE: Does the limit superior of every subsequence equals to the same measurable function imply convergence of the original sequence?
QUESTION [6 upvotes]: In an abstract measure space $(X, A, \mu)$, we consider a sequence of measurable functions $(f_n)_n$ with $f_n : X \to \mathbb{R}$ such that there exists some measurable function $f : X \to \mathbb{R}$, we have for every subsequence $\{n_k\}_k$ of $\{n\}$,
$$
     \limsup_{k \to \infty} f_{n_k} = f
     ~\mbox{a.e.}
$$
holds.
Then we ask if the following claim holds true
$$
      \lim_{n \to \infty} f_n = f
      ~\mbox{a.e.}
$$
Idea:
If the claim is false, I think maybe this example can help. $(X,A,\mu) = ([0,1], \mathcal{B}[0,1],m)$, $f = 1$, $f_n = 1_{A_n}$ with $A_n \in \mathcal{B}[0,1]$ and $m(A_n) = \frac{1}{2}$. The key is to construct $A_n$.

Since now the counterexample is found, I want to ask does the condition leads to convergence in measure induced by $f_n$, i.e. $\mu_n(A) := \mu(f_n \in A)$ for $A \in \mathcal{B}(\mathbb{R})$ satisfies $\lim_n \int_{\mathbb{R}} g(x) \mu_n(dx) = \int_{\mathbb{R}} g(x) \mu(dx)$ for any bounded continuous $g$ on $\mathbb{R}$?
if $\mu(X) = 1$, does the condition leads to convergence in probability of $f_n$, i.e. for any $\varepsilon$, does $\lim_n \mu(|f_n - f| \geq \varepsilon) = 0$?
The answer is no for two questions, if we take i.i.d. random variable with Bernoulli distribution.

REPLY [2 votes]: A counter example:
Let $f_n = 1_{A_n}$ with $m(A_n) = 1 - \frac{1}{n}$, $f = 1$.
$A_n$ is constructed in a way that it sweeps the interval $[0,1]$ for infinitly many times, i.e. $A_2 = [0, 1/2]$, $A_3 = [1/2, 5/6]$, $A_4 = [5/6, 1] \cup [1, 1/12]$, $A_5 = [1/12, 17/60]$.