In the book Banach Spaces of Vector-Valued Functions the authors present a demonstration for Proposition 1.6.4, pages 32-33:
Let $1\le p \le + \infty$, if $(\Omega,\Sigma,\mu)$ is a $\sigma$-finite purely atomic measure space then $L_p(\mu,X)$ is isometrically isomorphic to $\ell_p(X)$.
Let $(A_n)$ be a sequence of pairwise disjoint atoms of $(\Omega,\Sigma,\mu)$ whose union is $\Omega$. Then the map
\begin{align} L_p(\mu,X) &\longrightarrow \ell_p(X)\\ f=\sum_{n=1}^{\infty}\chi_{A_n}(\cdot)x_n &\longrightarrow (\mu(A_n)^{\frac{1}{p}}x_n)_n \end{align}
for $1\le p<\infty$, is a isometric isomorphism onto.
I am having difficulty understanding this map and consequently seeing that it is an isomorphism. $x_n$ is the point in $A_n$?
If $A$ is an atom then any measurable function $f$ on $A$ is a.e. constant on $A$. Proof: Consider $f^{-1}([\frac {i-1} 2,\frac i 2)$. These form a a partition of $A$. By definition of an atom it follows that $f$ takes values in $[\frac {i_1-1} 2,\frac {i_1} 2)$ for some $i_1$. Next look at $f^{-1}([\frac {i_1-1} {2^{2}},\frac {i_1} {2^{2}})$ and so on. You will see that $f$ takes values in a decreasing sequence of intervals $I_1,I_2,...$ with length of $I_n$ tending to $0$. There is a unique point $x$ in the intersection of these intervals and $f=x$ a.e.
Going back to the proof it is straightforward to compute the norm of $(\mu(A_n)^{1/n}x_n)_n$ in $\ell^{p}(X)$ and show that it is equal to norm of $f$.