Showing that a subspace of $L^p$ is closed I would like to prove that a particular subspace of $L^p([1,\infty[)$ (for some $p \in [1,\infty[$) is closed, but I'm not sure how to do it properly.

For any sequence $(x_n) \in \ell^p$, let define 
  $f_{(x_n)} : a \mapsto x_{\lfloor a \rfloor}$ (which is a "step" function, belonging to $L^p([1,\infty[)$).
  Then, I want to show that the subspace $E = \{f_{x} \mid x \in l^p \}$ is closed in $L^p([1,\infty[)$.

My idea was to take a sequence $f_{\,\stackrel{\rightarrow}{x}_n}$ in $E$ (where $\stackrel{\rightarrow}x_n = (x_{m,n})_{m \geq 1}$ is an element of $l^p$, for all $n ≥1$) that converges (for the $L^p$-norm) to some $g \in L^p$. Then I want to show that $g \in E$.
I think that I could use the fact that some subsequence of $(f_{\,\stackrel{\rightarrow}{x}_n})_{n ≥ 1}$ converges almost everywhere to $g$. But I find that this would be complicated ; anyway I don't see how to continue.
This is why I'm asking for some help. Thank you!
 A: In order to show the closeness we have to prove that if a sequence $(x_n)\in E \rightarrow x \in L^p \Rightarrow x \in E$.
Assume we have such sequence and $x \not\in E$, it means that $\exists z \in [1,\infty[$ such that $\forall \varepsilon > 0 \; x((z-\varepsilon,z+\varepsilon))$ is not constant whenever we change a set of measure zero.
But by construction $(x_n)$ is constant $\forall n$ in the above interval (if we are on a edge just take the left or right side). So 
$$\|x_n - x\|_{L^p(z-\varepsilon,z+\varepsilon)} \leq 
\text{ess}\!\!\!\sup_{(z-\varepsilon,z+\varepsilon)}|x_n-x|^p \times 2\varepsilon > \delta >0.$$
This is impossible since
$$\|x_n-x\|_{L^p}=\|x_n-x\|_{L^p(]1,\infty] \setminus (z \pm \varepsilon))} + \|x_n-x\|_{L^p(z \pm \varepsilon)} > \delta$$ and there is no convergence anymore.
A: Hint
Let $f_n$ a sequence of $L^p$ that converge to $f$. Since $x\longmapsto |x|^p$ is convex when $p\geq 1$, $$|f|^p=2^p\left|\frac{f-f_n+f_n}{2}\right|^p\leq 2^{p-1}(|f_n-f|^p+|f_n|^p).$$
