Discrete version of dominated convergence thm

Let $f_1,f_2,\ldots,g\colon\mathbb{Z}\rightarrow\mathbb{R}$ be functions such that $|f_N(n)|\leq g(n)$, $\sum_{n=-\infty}^\infty g(n)<\infty$, and $\lim_{N\rightarrow\infty}f_N(n)=f(n)$. Then show that $$\lim_{N\rightarrow\infty}\sum_{n=-\infty}^\infty f_N(n)=\sum_{n=-\infty}^\infty f(n).$$

This looks like the dominated convergence theorem, but how can we prove it directly?

Edit: As T. Bongers helpfully pointed out, this can be shown using the dominated convergence theorem. Is there a direct way to do it without the theorem?

Solution 1:

The direct proof would still follow the proof of DCT (with simplifications). It suffices to show that $$\liminf_{N\to\infty} \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{n\in\mathbb Z} f(n) \tag{1}$$ because by applying (1) to $-f_N$ and $-f$ you get the reverse inequality which completes the proof. In order to show (1), add $\sum_{n\in\mathbb Z} g(n)$ to both sides. This reduces (1) to the case when all functions are nonnegative. Given $\epsilon>0$, pick $K$ such that
$$ \sum_{|n|\le K} f(n)> \sum_{n\in\mathbb Z} f(n)-\epsilon$$ Due to pointwise convergence, for all sufficiently large $N$ you have $f_N(n) > f(n)-\epsilon/K$, for every $n=-K,\dots,K$. Hence, for such $N$ $$ \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{|n|\le K} f_N(n) >\sum_{|n|\le K} f(n)-3\epsilon > \sum_{n\in\mathbb Z} f(n)-4\epsilon$$ which proves (1).

Solution 2:

Since $\sum\limits_{n\in\mathbb{Z}} g(n)$ converges, for each $k$, $\sum\limits_{n\in\mathbb{Z}}f_k(n)$ converges. Furthermore, for any $\epsilon\gt0$, there is an $n_\epsilon$ so that $$ \sum_{|n|\gt n_\epsilon}g(n)\le\frac\epsilon2\tag1 $$ Thus, for any $\epsilon\gt0$, $$ \begin{align} \hspace{-9pt}\limsup_{k\to\infty}\left|\,\sum_{n\in\mathbb{Z}} f_k(n)-\sum_{n\in\mathbb{Z}}f(n)\,\right| &\le\lim_{k\to\infty}\sum_{|n|\le n_\epsilon}\!\!|f_k(n)-f(n)|+\limsup_{k\to\infty}\!\!\sum_{|n|\gt n_\epsilon}\!\!|f_k(n)-f(n)|\\ &\le0+\epsilon\tag2 \end{align} $$ Since $(2)$ is true for any $\epsilon\gt0$, $$ \lim_{k\to\infty}\sum_{n\in\mathbb{Z}}f_k(n)=\sum_{n\in\mathbb{Z}}f(n)\tag3 $$