Does $\int _{\Omega} \left | f_{n} \right |^p \rightarrow \infty $ implies $\left | f_{n}(x) \right | \rightarrow \infty $ pointwise?

Question

Let $f_{n}$ be sequence of Lebesgue integrable functions. $\Omega$ is open, bounded subset of $\mathbb{R}^n$.

We know that if $f_n$ converges to some limit in $L^p$, then we have a subsequence $ f_{n_{k}} $ which is bounded pointwise by a $L^p$ function. I don't know how it changes when the limit is infinity. Do we have a subsequence $ f_{n_{k}} $ that tends to infinity pointwise on some positive-measure subset of $\Omega$ (or entire $\Omega$)?

Thank you.

Answer 1

No. Consider $\Omega=(0, 1)$ and $f_n(x)=f(x)=|x|^\frac{-1}{p}$.

Re to comment. If you want the sequence to be in $L^p(0, 1)$, redefine $f_n$ as follows: $$ f_n(x)=\mathbb{1}_{\left[\frac1n, 1\right]}(x)\cdot f(x),$$ where $$\mathbb{1}_A(x)=\begin{cases} 1, & x\in A \\ 0, & x\notin A\end{cases}$$

Answer 2

Take $f_n$ to be $0$ on $[0,1]$, except on $[1-\frac{1}{n},1-\frac{1}{n+1}]$ where $f_n(x) = n(n+1)n^\frac{1}{p}$. So $$I_n^p = \int_0^1 (f_n(t))^p dt = n\,.$$ $I_n^p \to \infty$, while $f_n$ is $0$ (at least) on $[0,1-\frac{1}{n}[$.