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. 



Uniform boundedness in $L^1[0,1]$ implies finite $\limsup$ almost everywhere for a subsequence?
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 $\epsilon$.)


