Proving space is banach

Question

Let $$X=\left\{u\in L^1(\mu):\int_\Omega u~ d\mu=0\right\}$$ I want to show $(X,\|\cdot\|_{L^1})$ is banach.

Subspace of normed space is normed space so I just have to show convergence. Let $u_n$ be Cauchy in $X$. I want to show it has a limit in $X$

$$\|{u_n}\|_{L^1}= \int {\lim_n u_n}~d\mu =\int \lim_n |u_n|~d\mu=\lim_n\int|u_n|d\mu <\infty$$ since the $u_n$ are themselves in $L^1$ so the limit is in $L^1$. How do I show the sequence converges?

Answer 1

Prove that $X$ is closed (this follows from the fact that $u\mapsto\int_\Omega u\,\mathrm d\mu$ is continuous) and use the fact that a closed subspace of a complete space is again complete.

Answer 2

$L^{1}$ is complete, so you only have to show that $X$ is a closed subspace of $L^{1}$. Let $f_n \to f$ in $L^{1}$ norm and suppose $\int f_n d\mu=0$ for all $n$. Then $\int fd\mu=\lim \int f_nd\mu=0$ so $f \in X$.