L^1 convergence and limsup of convergent sequence

Question

I have to solve this exercise: let $f_n$ be a sequence of positive real function defined on a measure space $(X,M,\mu)$ such that $f_n\in L^1(\mu)$ $\forall n\in \mathbb{N}$ and $f_n$ is convergent in $L^1$-norm. Is it true that $\limsup_{n\to\infty}f_n$ is a.e. finite? If so, does it have to be Lebesgue-integrable?

I know that if $f_n\to f\in L^1(\mu)$ in $L^1$-norm there exists a subsequence $f_{n_k}$ pointwise convergent to the $L^1$-limit of $f_n$, but this does not exclude the existence of a divergent subsequence on a positive measure set, even if I think that the answer to the first question is yes. How can I prove it?

I thought about using Reverse Fatou's Lemma but it doesn't seem to help...

Answer 1

We endow the unit interval with Borel $\sigma$-algebra and Lebesgue measure. Denote for $n\geqslant 1$ and $0\leqslant k\leqslant 2^n-1$ the interval $I_{n,k}:=\left[k/2^n,(k+1)2^{-n}\right)$.

If $N$ is such that $2^n\leqslant N\lt 2^{n+1}$ for some $n\geqslant 1$, define $f_N(x):=n\mathbf 1_{\left(I_{n,N-2^n}\right)}(x)$.

In this way, $\lVert f_N\rVert_1=n2^{-n}\leqslant 2\log_2N/N$ hence $f_N\to 0$ in $\mathbb L^1$. However, for any $x\in [0,1)$, there exists an increasing sequence of integers $\left(t_n\right)_{k\geqslant 1}$ (depending on $x$) such that $f_{t_n}(x)\to +\infty$.

Indeed, let $x\in [0,1)$. For any $n\geqslant 1$, there is $k_n(x)$ such that $x\in I_{n,k_n(x)}$. Then take $t_n:=2^n+k_n(x)$. In this way, $f_{t_n}(x)=n$.

Answer 2

HINT If we let $$ g = \limsup_{n \rightarrow \infty} f_n,$$ then there is a subsequence $\{f_{n_k} \}$ of $\{ f_n \}$ such that $f_{n_k} \rightarrow g$ as $k \rightarrow \infty$.