Let $\Omega \subseteq \mathbb{R}^n$ be bounded and $p \in [1, \infty)$. Now let $(f_n)_{n \in \mathbb{N}} \subseteq L^p(\Omega)$ be a sequence and $f \in L^p(\Omega)$. Clearly, $$ \lVert f_n - f \rVert_{L^p(\Omega)} \overset{n \rightarrow \infty}{\longrightarrow} 0 \quad \implies \quad \int_\Omega \lvert f_n \rvert^p~\mathrm{d}x \overset{n \rightarrow \infty}{\longrightarrow} \int_\Omega \lvert f \rvert^p~\mathrm{d}x. $$
But I see no reason why $$ \lVert f_n - f \rVert_{L^p(\Omega)} \overset{n \rightarrow \infty}{\longrightarrow} 0 \quad \impliedby \quad \int_\Omega \lvert f_n \rvert^p~\mathrm{d}x \overset{n \rightarrow \infty}{\longrightarrow} \int_\Omega \lvert f \rvert^p~\mathrm{d}x $$ should hold. Unfortunately, I can't find a counterexample to disprove the latter. Can someone help me out and show me a counterexample for at least one combination of $\Omega$, $p$?
Note that integrals of non-negative functions are just non-negative numbers. Pick any non-negative convergent sequence $\{x_n\}_{n\in \mathbb N}$ and associate to each $x_n$ a function $f_n$ such that $\int_{\Omega }|f_n|^p\mathop{}\!d \mu=x_n$. Then there is no reason to be a relation between the $f_n$.
A simple counterexample could be setting $f_n:=\mathbf{1}_{[0,1+1/n]}$ for even $n$, $f_n:=\mathbf{1}_{[1,2+1/n]}$ for odd $n$ and $f:=\mathbf{1}_{[5,6]}$, then $$\int_{\mathbb{R}}|f_n|^p\mathop{}\!d \lambda =1+1/n\to 1=\int_{\mathbb{R}}|f|^p \mathop{}\!d \lambda,\quad \text{ as }n\to \infty $$ for any chosen $p\in [1,\infty )$ but $\|f_n-f\|_p^p=2+1/n$ for all $n$.