The completeness of ${L^p}({\mathbb R^n},B)$ Is ${L^p}({\mathbb R^n},B)$ complete for $p \ge 1$ where $B$ is a Banach space?
 A: *

*If $1\leq p<\infty$, let $\{f_n\}$ be a Cauchy sequence in $L^p(\Omega,B)$. We will show that there is a converging subsequence. We can assume that $\lVert f_n-f_m\rVert_{L^p(\Omega,B)}\leq 2^{-n}$ for $m>n$, after taking a subsequence, denoted in the same way. Put $A(n):=\left\{x\in\Omega\mid \lVert f_n(x)-f_{n+1}(x)\rVert_B\geq n^{-2}\right\}$. Then $\mathbf 1_{A(n)}n^{-2}\leq \lVert f_n(x)-f_{n+1}(x)\rVert_B$ for all $x$ hence 
$$\mu(A_n)^pn^{-2p}\leq \int_{\Omega}\lVert f_n(x)-f_{n+1}(x)\rVert_B^pd\mu(x)\leq 2^{-np}$$
so $\mu(A_n)\leq n^22^{-n}$. Put $B(n):=\bigcup_{m\geq n}A(n)$. The sequence $\{B(n)\}$ is decreasing and $\mu(B(n))\to 0$.
We define $N:=\bigcap_n B(n)$. If $x\notin N$, then for $n$ large enough we have $\lVert f_n(x)-f_{n+1}(x)\rVert_B\leq n^{-2}$, and since $B$ is complete, let $f(x):=\lim_n f_n(x)$. Put $f(x)=0$ if $x\in N$. Then we use Fatou lemma to get that $f\in L^p(\Omega,B)$ and $f_n\to f$ in $L^p(\Omega,B)$:
$$\int_{\Omega}\lVert f(x)\rVert^pd\mu(x)\leq \liminf_n\int_{\Omega}\lVert f_n(x)\rVert^pd\mu(x)<\infty$$
since a Cauchy sequence is bounded, and 
$$\int_{\Omega}\lVert f_n(x)-f(x)\rVert^pd\mu(x)\leq \liminf_m\int_{\Omega}\lVert f_n(x)-f_m(x)\rVert^pd\mu(x)\leq 2^{-np}.$$
This concludes the proof in the case $p$ finite. 

*If $p=+\infty$ and $\{f_n\}$ is a Cauchy sequence in $L^{\infty}(\Omega,B)$, we have for almost every $x$ $\lVert f_n(x)-f_m(x)\rVert_B\leq \lVert f_n-f_m\rVert_{\infty}$. We use the completeness of $B$ to get a limit $f(x)$ for these $x$, and we put $f(x)=0$ for the other $x$. We get that $f\in L^{\infty}(\Omega,B)$ and $f_n\to f$ in $L^{\infty}(\Omega,B)$.
