Let \((g_k)\subset L^1(\RR^n)\) and suppose \[ \sum_{k=1}^\infty\norm{g_k}_1<\infty. \]
Show that \(\sum_{k=1}^\infty g_k\) converges almost everywhere to some \(g\in L^1\).
Show that the partial sums converge to \(g\) in \(L^1\).
Solution
1. The function \(\sum_{k=1}^\infty\abs{g_k}\) is integrable and finite almost everywhere.
By Tonelli's theorem, \[ \int_{\RR^n}\sum_{k=1}^\infty\abs{g_k(x)}\,dx =\sum_{k=1}^\infty\norm{g_k}_1<\infty. \] Hence \(\sum_{k=1}^\infty\abs{g_k(x)}<\infty\) for almost every \(x\).
2. The series \(\sum_{k=1}^\infty g_k\) converges almost everywhere to some \(g\in L^1(\RR^n)\).
By step 1, the series converges absolutely for almost every \(x\). Define \(g(x)\) to be its sum there and set \(g(x)=0\) on the null exceptional set. Then \[ \abs{g(x)}\le\sum_{k=1}^\infty\abs{g_k(x)} \] almost everywhere, so step 1 gives \(g\in L^1(\RR^n)\).
3. If \(S_N=\sum_{k=1}^N g_k\), then\[ \norm{g-S_N}_1\longrightarrow0. \]
Almost everywhere, \[ \abs{g-S_N}\le\sum_{k>N}\abs{g_k}. \] Therefore \[ \norm{g-S_N}_1 \le\sum_{k>N}\norm{g_k}_1\longrightarrow0, \] because the series of norms converges.
4. Q.E.D.
Step 2 proves part (a), and step 3 proves part (b).