Let \((X,\mcm,\mu)\) be a measure space and let \(1\leq p\leq\infty\). Then \(L^p(X,\mu)\), whose elements are classes of functions modulo equality \(\mu\)-almost everywhere, with the norm \(\norm{\cdot}_p\), is a Banach space.
First let \(1\leq p<\infty\), and let \((f_n)\) be Cauchy in \(L^p\). Choose a subsequence \((f_{n_k})\) such that \[ \norm{f_{n_{k+1}}-f_{n_k}}_p\leq 2^{-k} \qquad(k\geq1). \] Set \[ g_k\coloneqq\abs{f_{n_{k+1}}-f_{n_k}}, \qquad S_N\coloneqq\sum_{k=1}^N g_k. \] Minkowski’s inequality gives \[ \norm{S_N}_p\leq\sum_{k=1}^N2^{-k}\leq1. \] Since \(S_N\uparrow S\coloneqq\sum_{k\geq1}g_k\), the monotone convergence theorem yields \[ \int_X S^p\dmu =\lim_{N\to\infty}\int_X S_N^p\dmu \leq1. \] Thus \(S<\infty\) almost everywhere. Outside a null set, the series \[ \sum_{k\geq1}\big(f_{n_{k+1}}(x)-f_{n_k}(x)\big) \] converges absolutely; there, define \(f(x)\) so that this sum equals \(f(x)-f_{n_1}(x)\), and put \(f\coloneqq0\) on the exceptional null set.
For every \(k\) and \(N>k\), \[ \abs{f_{n_N}-f_{n_k}} \leq\sum_{j=k}^{N-1}g_j. \] Letting \(N\to\infty\) and applying Fatou’s lemma together with Minkowski’s inequality gives \[ \norm{f-f_{n_k}}_p \leq\sum_{j=k}^{\infty}\norm{g_j}_p \leq\sum_{j=k}^{\infty}2^{-j} =2^{1-k}. \] Hence \(f_{n_k}\to f\) in \(L^p\). Given \(\varepsilon>0\), choose \(k\) with \(\norm{f_{n_k}-f}_p<\varepsilon/2\) and \(n_k\geq N\), where \(N\) is such that \(\norm{f_n-f_m}_p<\varepsilon/2\) for all \(m,n\geq N\). Then \(\norm{f_n-f}_p<\varepsilon\) for all \(n\geq N\), so \(f_n\to f\) in \(L^p\).
Now let \(p=\infty\), and again choose a subsequence with \[ \norm{f_{n_{k+1}}-f_{n_k}}_\infty\leq2^{-k}. \] For each \(k\) there is a null set \(Z_k\) outside which \[ \abs{f_{n_{k+1}}-f_{n_k}}\leq2^{-k}. \] Outside the null set \(Z\coloneqq\bigcup_kZ_k\), the series of successive differences converges uniformly by the Weierstrass \(M\)-test. Hence \((f_{n_k})\) converges uniformly on \(X\setminus Z\) to a bounded measurable function \(f\), extended by \(0\) on \(Z\), and \[ \norm{f-f_{n_k}}_\infty\leq2^{1-k}. \] As in the case \(p<\infty\), since \((f_n)\) is Cauchy, \(f_n\to f\) in \(L^\infty\).