Pointwise convergence at a countable number of points Let $(f_n)_n$ and $f$ be real-valued functions. Let $D$ be a countable collection of points on the real line. I have that
$$f_n(x_i) \to f(x_i) \tag{1}$$
for every $x_i \in D$.
Without imposing any conditions on $(f_n)$ or $f$, does it hold that $(1)$ is true if and only if
$$\sum_{i=1}^{\infty}\lvert f_n(x_i) - f(x_i)\rvert 2^{-i} \to 0$$
as $n\to\infty$ ?
If yes, how does one show this? If not, what does one need to impose on the functions?

Look at $x_i=i$ and at some sequence of functions which satisfy $f_n(x_n) = 2^n$ but $f_n(x_i)=0$ whenever $i\neq n$. Then $f_n\rightarrow 0$ pointwise, but the sum you are looking at is always equal to $1$.
(So the answer to your question is 'no'. I know this is not a full answer to your question, but it should answer parts of it)
(One more thought: do you know some measure theory? 
If yes, consider the measure $\mu$ which is defined by $\mu(\{x_i\}) = 2^{-i}$ and $\mu(A) = 0$ if $A\cap D$ is empty. Then the question you are asking is whether pointwise convergence (on $D$) is equvivalent to $\int |f_n-f| d\mu \rightarrow 0$. This looks like a simple reformulation, but now you have all the theorems from Lebesgue integration theory available).
