I have a question about the proof from the book ''Banach Spaces of Continuous Functions as Dual Spaces'', stated as Proposition 2.4.8, that is
The Banach space $\ell^1$ is isometrically isomorphic to a $1-$complemented subspace of $L^1([0,1])$.
Proof starts with the following: let $\{ I_n: n \in \mathbb{N} \}$ be a family of pairwise-disjoint, closed intervals in $[0,1]$ and for each $n \in \mathbb{N}$, let $\chi_n$ be the characteristic function of $I_n$, $l_n$ the length of $I_n$, and $f_n := \frac{\chi_n}{l_n}$, so that $\| f_n \|_1 = 1$. The author then denotes $E$ as the closure of the linear span of $\{f_n: n \in \mathbb{N} \}$. The idea is that $E$ is isometrically isomorphic to $\ell^1$ via $T: \ell_1 \rightarrow E$ defined by $$T((a_n)_{n=1}^\infty) := \sum_{n=1}^\infty a_n f_n$$ and finally author shows that the map $P : L^1[0,1] \rightarrow L^1[0,1]$ defined by $$P(f):=\sum_{n=1}^\infty f_n \int_0^1 f(t)dt $$ is the bounded projection onto $E$.
The part that I don't understand is the construction of functions $f_n$. Since we are picking disjoint closed intervals in $[0,1]$ (of positive length, since we divide by it in the construction of $f_n$), we have to have a finite amount of non-empty intervals inside this family (as we know from Is $[0,1]$ a countable disjoint union of closed sets? for example), so does it mean that we have finitely many non-zero $f_1,f_2,...f_n$ and for $k>n$ we have $f_k=0$? I don't see how this can work since then this would imply that $E$ is a finitely dimensional subspace of $L^1[0,1]$ that is isometrically isomorphic to $\ell^1$, which is infinitely dimensional? This doesn't make much sense, but I don't see how we could prove that $T$ is onto $E$ without the ''finite'' part, since if we take $f \in E$, so $f = \sum_{n=1}^\infty a_n f_n$ for some scalars $a_n$, we have to have that only finitely many out of $a_n$ are non-zero, to ensure that $(a_n)_{n=1}^\infty \in \ell^1$.
Could anyone point out the mistake in what I wrote? I feel pretty lost with this construction, and it doesn't seem that difficult.
Thanks!
