Counting measure in integral. Let $\mu$ counting measure on $X$ countable set.
This is correct?
$\int_{X}|f|^pd\mu=\int_{\bigcup_{k\in\mathbb{N}} x_k} |f|^pd\mu=\sum_{k\in\mathbb{N}}\int_{x_k}|f|^pd\mu=\sum_{k\in\mathbb{N}}\int |f|^p\mathcal{X}_{x_k}d\mu=\sum_{k\in\mathbb{N}} |f(x_k)|^p\int \mathcal{X}_{x_k}=\sum_{k\in\mathbb{N}} |f(x_)|^p\mu(x_k)=\sum_{k\in\mathbb{N}}|f(x_k)|^p$
Therefore ${L(X)}^p={l(X)}^p$

The equalities are in principle correct provided that $X$ is not finite. 
Not completely correct is the notation that is practicized. 
Every $x_k$ in the equalities should be replaced by $\{x_k\}$.

Further IMV it is a bit overdone.
If $\mu$ denotes the counting measure and $X$ is countable then you can write immediately:$$\int_X|f|^pd\mu=\sum_{x\in X}|f(x)|^p$$
There is no need to go for $\mathbb N$ as index set. It is even wrong if $X$ is a finite set.

Where it concerns $L(X)^p=l(X)^p$ see the comment of Kavi to your question.
