For an arbitrary set $I$, not necessarily countable, is the space $\ell^1(I)$ consist of functions on $I$ whose support is always countable? Here "support" doesn't refer to the closure, but only those elements in $i\in I$ such that $f(i)\neq 0$, where $f\in \ell^1(I)$.
This is because, for an uncountable set $I$, the norm $$\|f\|_1:=\sup_{F\in \mathcal{F}}\left\{\sum_{j\in F} |f(i_j)|\right\}< +\infty,$$ where $\mathcal{F}$ is the set/family of all finite subsets of $I$, forces fucntion $f$ only takes nonzero values on at most countably many points $i_j\in I$.
Is that correct? I know $\ell^1(I)$ is equivalent to $L^1(I)$ through the counting measure.
Yes, if $f\in \ell^1(I)$, then $f$ has countable support. Indeed, note that
$$\{i \in I: f(i)\ne 0\}= \bigcup_{n=1}^\infty \{i\in I: |f(i)|> n^{-1}\}.$$ The set $\{i \in I: |f(i)| > n^{-1}\}$ is finite for every $n$, otherwise you could pick an infinite subset $I_0$ with the property that $|f(i)| > n^{-1}$ for all $i\in I_0$ and then $$\|f\|_1=\sum_{i \in I}|f(i)|\ge \sum_{i \in I_0}|f(i)| \ge \sum_{i \in I_0} n^{-1}=\infty.$$ Hence, the support of $f$ is a countable union of finite sets, and thus a countable set.