Almost everywhere convergence can lose mass by escape or concentration, so $L^p$ convergence need not follow.Suggest a slogan
For every \(1\le p\le\infty\) there is a sequence \((f_k)_{k\geq1}\) in \(L^p(\RR)\) with \(f_k\to0\) almost everywhere but \(\norm{f_k}_p\not\to0\).
Escape to infinity. Let \(f_k\coloneqq\chi_{[k, k+1]}\). For every \(x\in\RR\), \(f_k(x)=0\) once \(k>x\), so \(f_k\to0\) pointwise everywhere. But \(\norm{f_k}_p = 1\) for all \(k\) and all \(1\le p\le\infty\). The convergence is not uniform, since \(\sup_x\abs{f_k(x)}=1\).
Concentration. Let \(f_k\coloneqq k \chi_{[0, 1/k]}\). For \(x\neq0\), \(f_k(x)=0\) whenever \(k>1/\abs{x}\), so \(f_k\to0\) almost everywhere; at \(x=0\), \(f_k(0)=k\to\infty\). Here \(\norm{f_k}_p=k^{1-1/p}\geq1\) for \(1\le p<\infty\) and \(\norm{f_k}_\infty = k \to \infty\).