Completeness of $(BMO(\Bbb R^n),||\cdot||_{\ast})$ Recall that
$$
BMO(\Bbb R^n):=\left\{f\in L_{loc}^1(\Bbb R^n)\;\mbox{modulo constant functions, such that}\\
\forall B\subseteq\Bbb R^n\;\mbox{ball}, \exists\alpha(B)\in\Bbb R\;\;\mbox{such that}\\
\frac1{|B|}\int_B|f(x)-\alpha(B)|\,dx\le C\;\;\;\;\exists C>0\right\}
$$
Observe that $C$ is uniform in $B$.
Given $f\in BMO$, the infimum of such $C$'s is the $BMO$ norm of $f$, denoted by $||f||_*$.
In particular, elements of $BMO$ are not functions, they are classes of functions.
I have to prove that $(BMO(\Bbb R^n),||\cdot||_{\ast})$ is complete.
So let's take a Cauchy sequence $\{f_n\}_{n\ge1}\subseteq BMO(\Bbb R^n)$; thus
$$
||f_n-f_m||_*\to0\;\;\;\mbox{as}\;\;\;n,m\to+\infty.
$$
So we can extract a subsequence $\{f_{n_k}\}_{k\ge1}\subseteq \{f_n\}_{n\ge1}$ such that, given $\epsilon>0\;\;\;\exists N_{\epsilon}\ge1$, such that
$$
||f_{n_k}-f_{n_1}||_*<\epsilon 2^{-k}\;\;\;\forall k\ge1
$$
from which clearly
$$
||f_{n_k}-f_{n_1}||_*\to0\;\;\;\mbox{as}\;\;k\to+\infty
$$
i.e. $\{f_n\}_n$ admits a converging subsequence, thus $\{f_n\}_n$ itself converges, so we have done.
But this doesn't sounds good.
The fact we are dealing with classes and not with functions must play some role, but I can't understand what is.
Can you tell me if it's right or not please?
Many thanks
 A: It is well known that if $f \in \text{BMO}(\mathbb R^n)$, then for any $\epsilon > 0$ there exists a ball $B$ with center $x_0$ and radius $R$ such that
\begin{equation}\label{3}
R^\epsilon \int_{\mathbb R^n} \frac{|f(x) - f(B)|}{(R + |x - x_0|)^{n + \epsilon}} \, dx \leq C_{n,\epsilon} \|f\|_*\;.
\end{equation}
We can deduce from this theorem that any $\text{BMO}$-Cauchy sequence is Cauchy in $L^1$ on every compact subset of $\Bbb R^n$.
From this we will deduce that every Cauchy sequence converges.
\newline
First of all, setting $\int_B^*:=\frac1{|B|}\int_B$  let's define
$$
\text{quasi-BMO}(\Bbb R^n):=\left\{f\in L_{loc}^1(\Bbb R^n)\;:\;
\forall B\subseteq\Bbb R^n\;\mbox{ball}\\
\int_B^*|f(x)-f(B)|\,dx\le C\;\;\;\;\exists C>0\right\}
$$%\end{align*}
Given $f\in\text{quasi-BMO}$, let's define then
$$
\|f\|_q:=\sup_B\fint_B|f(x)-f(B)|\,dx
$$
the problem is that $\|\cdot\|_q$ is actually a seminorm, that is $\|f\|_q = 0$ does not only occur when $f = 0$ but actually for all constants.
$\text{BMO}$ space is built defining an equivalence relation on $\text{quasi-BMO}$ in order to making it properly a normed space.
So let $f \sim g$ if and only if $f - g$ is a constant. The normed space we call $\text{BMO}$ is thus the quotient space $\text{quasi-BMO}/\{\mbox{const. funct.}\}$ and the projection $\pi:\text{quasi-BMO}\to\text{BMO}$ that sends every function to its equivalence class, i.e. $\pi(f):=[f]$, is linear and continuous.
Also, we have a norm on this space $\|[f]\|_* = \inf_{c \in \mathbf{R}} \|f + c\|_q$. If we want to show completeness, we would have to build a function where a Cauchy sequence converges to but this doesn't seem smart. We rather use the following:
$\textbf{Theorem:}$ A normed space $(X,\|\cdot\|)$ is complete iff $\sum_m x_m$ converges in the norm for every sequence $\{x_m\}_m\subseteq X$ such that $\sum_m \|x_m\|<+\infty$.
First of all, note that $\|[f]\|_*=\|f\|_q$ for all $f\in\text{quasi-BMO} $:
\begin{align*}
\|[f]\|_*
&= \inf_{c\in\R} \|f + c\|_q\\
&=\inf_{c\in\R}\sup_B\fint_B\left|f(x)+c-\fint_B(f(y)+c)dy\right|\,dx\\
&=\inf_{c\in\R}\sup_B\fint_B|f(x)-f(B)|\,dx\\
&=\sup_B\fint_B|f(x)-f(B)|\,dx\\
&=\|f\|_q
\end{align*}
Take now a sequence of classes of functions $\{[f_m]\}_m\subseteq\text{BMO}$
such that $\sum_m\|[f_m]\|_*<+\infty$.
So we have a sequence of functions $\{f_m\}_m\subseteq\text{quasi-BMO}$ such that  $\sum_m\|f_m\|_q<+\infty$.
Now let $B$ be a closed ball with radius $R$ centered in $x_0\in\Bbb R^n$.
Then 
\begin{align*}
\frac{R^{\epsilon}}{(2R)^{n+\epsilon}}\|f_m\|_{L^1(B)}
&=R^{\epsilon}\int_B\frac{|f_m(x)|}{(2R)^{n+\epsilon}}\,dx\\
&\le R^{\epsilon}\int_B\frac{|f_m(x)-f_m(B)|}{(2R)^{n+\epsilon}}\,dx+\frac{R^{\epsilon}}{|B|(2R)^{n+\epsilon}}\overbrace{\int_B|f_m(x)|\,dx}^{\|f_m\|_{L^1(B)}}\\
\end{align*}
from which (we take wlog $|B|>1$)
\begin{align*}
\left(1-\frac1{|B|}\right)\left(\frac{R^{\epsilon}}{(2R)^{n+\epsilon}}\right)\|f_m\|_{L^1(B)}
&\le R^{\epsilon}\int_B\frac{|f_m(x)-f_m(B)|}{(2R)^{n+\epsilon}}\,dx\\
&\le R^\epsilon \int_{B} \frac{|f_m(x)-f_m(B)|}{(R + |x-x_0|)^{n + \epsilon}} \, dx\\
%&=R^\epsilon \int_{B} \frac{|f_m(x)-f_m(B)|}{(R + |x |)^{n + \epsilon}} \, dx\\
&\le R^\epsilon \int_{\Bbb R^n} \frac{|f_m(x)-f_m(B)|}{(R + |x-x_0|)^{n + \epsilon}} \, dx\;\tag{see (\ref3)}\\
&\le C_{n,\epsilon} \|f_m\|_q\;\;.\\
%&\le C_{n,\epsilon} \|[f_m]\|_*\\
\end{align*}
Take now a compact subset $K\Subset\Bbb R^n$; so there exist $R>0$, $x_0$ such that $K\subseteq B$ and $|B|>1$; from this we deduce that
$$
\sum_m\|f_m\|_{L^1(K)}\le
\sum_m\|f_m\|_{L^1(B)}\le
C\sum_m\|f_m\|_q<+\infty
$$
so being $L^1(K)$ complete, by the Theorem stated above we have that $\sum_m f_m$ converges in $L^1(K)$, to a function we call $F_K$, i.e. $\|\sum_m f_m-F_K\|_{L^1(K)}\to0$ as $m\to+\infty$.\
Define $F:\Bbb R^n\to\Bbb R$ to be $F_K$ on every compact subset $K$. Let's see $F$ is well defined.\
If $K_1,K_2\Subset\Bbb R^n$ are such that $K_1\cap K_2\neq\emptyset$, let's show that
$$
{{F_{K_1}}_|}_{K_1\cap K_2}\equiv
{{F_{K_2}}_|}_{K_1\cap K_2}.
$$
Clearly $\|\sum_m f_m-{F_K}_1\|_{L^1(K_1\cap K_2)}\le\|\sum _m f_m-{F_K}_1\|_{L^1(K_1)}\to0$; similarly  $\|\sum_m f_m-{F_K}_2\|_{L^1(K_1\cap K_2)}\to0$ thus $\sum_m f_m$ converges to both $F_{K_1}$ and $F_{K_2}$ in $L^1(K_1\cap K_2)$, so (possibly passing to a subsequence) $\sum_m f_m$ converges pointwise, a.e. on $K_1\cap K_2$, to both $F_{K_1}$ and $F_{K_2}$ which hence must here coincide, as wanted.\
So we have our $F\in L_{loc}^1$ which is well defined and $\sum_m f_m$ converges to $F$ in $L_{loc}^1(\Bbb R^n)$.\
Let's prove now that $\sum_m f_m$ converges to $F$ in $\text{quasi-BMO}$:
\begin{align*}
\fint_B \left|\sum_{m=m_0}^N f_m(x)-F(x)-\left(\sum_{m=m_0}^N f_m-F\right)(B)\right|\,dx
&\le\fint_B \left|\sum_{m=m_0}^N f_m(x)-F(x)\right|\,dx+\left|\left(\sum_{m=m_0}^N f_m-F\right)(B)\right|\\
&\le\frac2{|B|}\int_B \left|\sum_{m=m_0}^N f_m(x)-F(x)\right|\,dx\\
&=\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}
\end{align*}
from which we get
\begin{align*}
\left\|\sum_{m=m_0}^N f_m-F\right\|_q
\le\sup_B\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}
\end{align*}
thus
\begin{align*}
\lim_N\left\|\sum_{m=m_0}^N f_m-F\right\|_q
&\le\lim_N\sup_B\frac2{|B|}\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}\\
&=\sup_B\frac2{|B|}\lim_N\left\|\sum_{m=m_0}^N f_m-F_B\right\|_{L^1(B)}=0
\end{align*}
I know that the interchange between $\lim$ and $\sup$ should be checked, but it seems reasonable to accept it.\
Thus $\sum_m f_m$ converges to $F$ in $\text{quasi-BMO}$ and since
\begin{align*}
\left\|\sum_{m=m_0}^N f_m-F\right\|_q
&=\left\|\pi\left(\sum_{m=m_0}^N f_m-F\right)\right\|_* \tag{$\pi$ is linear}\\
&=\left\|\sum_{m=m_0}^N [f_m]-[F]\right\|_*
\end{align*}
we can conclude that the given sequence $\{[f_m]\}_m\subset \text{BMO}$ taken such that $\sum_m\|[f_m]\|_m<+\infty$, is then such that $\sum_m[f_m]$ converges in $\text{BMO}$, which is thus complete by the Theorem above. 
