Suppose we have a sequence of functions $f_n : \mathbb{N} \to [0,1]$ such that $$ |f_n|_{1} := \sum_{i=1}^{\infty} |f_n(i)| = A $$ for each $n$, where $A > 0$ is given. Suppose that for each $i$ we have $f_n(i) \to g(i)$ as $n\to\infty$, where $g : \mathbb{N} \to [0,1]$ and $|g|_1 = A$.
How do I prove that $|g - f_n|_1 \to 0$ as $n \to \infty$?
What I have tried: define $$ s_{n,m} := \sum_{i=1}^{m} |f_n(i) - g(i)| $$. Then $s_{n,m}$ has a global upper bound of $2A$ for each $n,m$. With $n$ fixed, since $s_{n,m}$ is monotone increasing in $m$, $c_n = \lim_{m \to \infty}s_{n,m}$ is well defined for each $n$. Moreover, for each fixed $m$ we can exchange limit and finite sum to have $\lim_{n \to \infty}s_{n,m} = 0$.
What I need to show now is that $\lim_{n\to\infty}c_n = 0$. So choose $\eta > 0$ and we need to find $N$ such that $|c_n| < \eta$ for all $n > N$. Now we see that
\begin{align*} |c_n| = \sum_{i=1}^{M_n} |f_n(i) - g(i)| + \sum_{i=M_n + 1}^{\infty} |f_n(i) - g(i)| \end{align*}
I can choose $M_n$ to make the second summation strictly less than $\frac{\eta}{2}$ since $c_n$ is finite. But I do not know how to deal with the first summation.
Any hints comments or solutions are appreciated. I am open to trying a different approach as well.
This is just a minor modification of what is known as Scheffe's Lemma.
$\sum_i (g(i)-f_n(i))^{+}\to 0$ by DCT (applied to counting measure) since $0 \le (g(i)-f_n(i))^{+}\le g(i)$ and $\sum g(i)<\infty$. Also, $\sum_i (g(i)-f_n(i))= 0$. Since $x^{-}=-x+x^{+}$ we get $\sum_i (g(i)-f_n(i))^{-}\to 0$. Adding $\sum_i (g(i)-f_n(i))^{+}\to 0$ and $\sum_i (g(i)-f_n(i))^{-}\to 0$ w get $\sum_i |g(i)-f_n(i)|\to 0$.
Proving that $\sum_i (g(i)-f_n(i))^{+}\to 0$ by DCT without DCT: Let $\epsilon >0$. There exists $N$ such that $\sum\limits_{i=N+1}^{\infty} (g(i)-f_n(i))^{+}\le \sum\limits_{i=N+1}^{\infty} g(i)<\epsilon /2$. Now $\sum\limits_{i=1}^{N}(g(i)-f_n(i))^{+} <\epsilon /2$ for $n$ sufficiently large and adding these two we get $\sum\limits_{i=1}^{\infty} (g(i)-f_n(i))^{+}<\epsilon$ for $n$ sufficiently large.