Let \((X,\mcm,\mu)\) be a measure space, let \(1\leq p<\infty\), and let \(f_n, f\colon X\to\CC\) be measurable.
If \(f_n\to f\) uniformly on \(X\), then \(f_n\to f\) pointwise on \(X\); if \(f_n\to f\) pointwise on \(X\), then \(f_n\to f\) almost everywhere.
If \(\mu(X)<\infty\), \(f_n, f\in L^p(\mu)\), and \(f_n\to f\) uniformly on \(X\), then \(f_n\to f\) in \(L^p(\mu)\).
If \(f_n\to f\) in \(L^p(\mu)\), then some subsequence \((f_{n_k})\) converges to \(f\) almost everywhere.
Without \(\mu(X)<\infty\), uniform convergence does not imply \(L^1\) convergence: \(\frac1n\chi_{[0,n]}\to 0\) uniformly on \(\RR\) with \(\int_\RR\frac1n\chi_{[0,n]} = 1\) (FE-3RPMC). Almost everywhere convergence does not imply \(L^1\) convergence: \(n\chi_{[0,1/n]}\to 0\) almost everywhere on \(\RR\) with integral \(1\) (FE-VT5N3).
\(L^1\) convergence does not imply almost everywhere convergence. Write each \(n\geq 1\) uniquely as \(n = 2^k + j\) with \(k\geq 0\) and \(0\leq j<2^k\), and let \(f_n\coloneqq\chi_{[j2^{-k},\,(j+1)2^{-k}]}\) on \([0,1]\). Then \(\norm{f_n}_1 = 2^{-k}\to 0\), but for every \(x\in[0,1]\) and every \(k\) some \(f_n\) with \(2^k\leq n<2^{k+1}\) has \(f_n(x) = 1\), and for \(k\geq 2\) another has \(f_n(x) = 0\), so \((f_n(x))\) diverges at every \(x\in[0,1]\).