Completeness of BMO without duality to $H^1$ for $f \in L^1_{loc}$, let the sharp-function be defined as
$f^\sharp(x) := \sup_{B \in x} |B|^{-1} \int_B |f(y) - |B|^{-1} \int_B f(z) dz | dy$
We define the space $BMO \subset L^1_{loc}$ via
$BMO := \{ f \in L^1_{loc} \;\;\;s.t.\;\;\; f^\sharp \in L^{\infty}  \}$
and the BMO-(semi)norm on BMO as
$\Vert f \Vert_{BMO} = \Vert f^\sharp \Vert_{\infty}$
The completeness of BMO can be infered from the fact that it is exactly the dual space of the Hardy space $H^1$ with operator norm.
Is there a more direct way to show completeness of BMO?
 A: Yes.
We have the following theorem:
Let $f \in \text{BMO}(\mathbb R^n)$. Then for any $\epsilon > 0$ there exists a ball $B$ with center $x_0$ and radius $R$ such that
$$R^\epsilon \int_{\mathbb R^n} \frac{|f(x) - \text{Avg}_B f|}{(R + |x - x_0|)^{n + \epsilon}} \, dx \leq C \|f\|_\text{BMO}$$
where $C$ only depends on $n$ and $\epsilon$. (this is just estimating stuff)
So from this theorem we can deduce that any $\text{BMO}$-Cauchy sequence is Cauchy in $L^1$ on every compact set. From this we can deduce that every Cauchy sequence converges.
Edit: (A slight extension) My apologies for the two year delay. First remark that the $\text{BMO}$ norm is actually a seminorm, that is $\|x\| = 0$ does not only occur when $x = 0$ but actually for all constants. This should be obvious, otherwise you better go check out a book on integration theory. Let us define an equivalence relation on this space to the end of making it a true normed space. That is let $f \sim g$ if and only if $f - g$ is a constant. If we now make the quotient space $\text{BMO}/\{1\}$ this is a normed space and the map that sends every function to its equivalence class is continuous. Also, we have a norm on this space $\|[f]\| = \inf_{c \in \mathbf{R}} \|f + c\|$. If we want to show completeness, we would have to cook up a function where a Cauchy sequence converges to. That sucks, so let us not do that but use a corollary.

Theorem: Let $X$ be a normed space and let $(x_n)$ be a sequence in $X$ such that $$\sum_n \|x_n\|$$ converges. Then $X$ is a Banach space if and only if $$\sum x_n$$ converges in the norm.

Cool. Let us use this. First of all, remark that the induced norm gives for all non-constant functions the normal $\text{BMO}$ seminorm and for the constants just the constant. As constant sequences are pretty much boring and certainly convergent we could as well not care about sequences that are "eventually constant". So let us consider the functions $g_n = f_n - \langle f_n \rangle$ as the important ones. This is cool, right? Take a sequence $(g_n)$ of such functions such that the above sum converges. Now let $B$ be a closed ball with radius $R$ and center $x_0$. Then we have,
$$R^\epsilon \int_{B} \frac{|g_n(x)|}{(R + |x - x_0|)^{n + \epsilon}} \, dx \lesssim \|g_n [B]\|_\text{BMO} \lesssim \|g_n\|_\text{BMO}.$$
This means that we have for all $\epsilon > 0$ there is $R > 0$ and an $x_0$ such that
$$R^\epsilon \sum_n \int_{B} \frac{|g_n(x)|}{(R + |x - x_0|)^{n + \epsilon}} \, dx$$
converges. Now note that $R \leqslant R + |x - x_0| \leqslant 2R$ hence, we get that
$$\frac1{R^d} \sum_n \|g_n\|_{L^1(B)}$$
converges and so also $g_n$ converges in $L^1(B)$ to a function $g_B$. Now you might be afraid that these $g_B$'s might be too different for different $B$'s, but not that if you enlarge the $B$, the "$g$" should at least coincide on the intersection of the balls.
Remark: I am too tired now, I will edit more later on. But now the idea would be to use this limit for each $B_k$ and pick a nice sequence of $B_k$'s. If the BMO sequence were to converge one would expect that the limit should be the same as the $L^1$ version (were it restricted). My idea would be to decompose $\mathbf{R}^d)$ into a sequences of shells. You start with $B$ and blow it up two times. Do this again with the new ball and so on. Then make this a disjoint sequence and build your $g$ from a sum of indicators of these shells and then multiply by the function you get from the $L^1(B_n)$. Show this is in BMO and that is is the limit you want. Cheers.
