Given a sequence of functions $f_k \in L^1([0,1])$ such that $||f_k||_{L^1(0,1)}\leq C$.
Is there a subsequence $\{k_l\,|\,l\in \mathbb N\}\subseteq \mathbb{N}$ such that for $\mathcal{L}^1$-almost every $x\in (0,1)$ it holds: $\limsup_{k_l\in \mathbb N} |f_{k_l}(x)| < \infty$?
Although the statement is quite easy to understand I was neither able to produce a counterexample nor a prove.
In general, this is not possible. I'll use the following fact:
Given $\delta, \epsilon >0$, we can find subsets $E_n\subset [0,1]$ with $|E_n|<\delta$ so that $\left|\bigcup E_{n_j} \right| > 1-\epsilon$ for any selection of infinitely many of these sets.
We can do this explicitly, as follows: Let $E_1=[0,\delta]$, and take disjoint translates of this interval to build the first generation of sets. The second generation will consist of translates of $E_n = \bigcup_j [j\delta, j\delta+\delta^2]$, and then the sets from the third generation will be unions of $1/\delta^2$ intervals of length $\delta^3$ each etc.
Suppose we have already selected $N$ sets. Then a set from a later generation will mainly be in the gaps of those sets, so the measure of the union will be $\gtrsim \delta +(1-\delta)\delta + (1-\delta)^2\delta + \ldots$.
Now start defining $f_n$'s by declaring $f_n=1$ on sets $E^{(1)}_n$ of this type, with $\delta_1=\epsilon_1=2^{-1}$. After a while, throw in a second series of sets $E^{(2)}_n$, with $\delta_2=\epsilon_2=2^{-2}$, and set $f_n=2$ on these. Continue in this way.
By construction, $\limsup f_{n_j}\ge N$ on a set of measure $\ge 1-2^{-N}$ for all $N$ and for any subsequence.
(Notice that $\limsup g_n <\infty$ a.e. would imply that $g_n(x)\le A$ for all $n\ge N_0$ and $x\notin B$, $|B|<\epsilon$, for some $A, N_0, B$, all depending on $\epsilon$.)
