Why $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$? 
We also deduce that if $f\in BMO$ is real-valued, and $f^{(k)}$ is the
truncation of $f$ defined by $f^{(k)}(x)=f(x)$ if $\mid f(x)\mid\leq
 k;f^{(k)}(x)=k$ if $f(x)>k$; and $f^{(k)}(x)=-k$, if $f(x)<-k$, then
$\{f^{(k)}\}$ is a sequence of bounded BMO functions so that $\mid
 f^{(k)}\mid\leq \mid f\mid $ for all $k,f^{(k)}\to f$ for a.e $x$ as
$k\to\infty$, and hence $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$ as
$k\to \infty$.
If $f$ is complex-valued, one may apply this to both the real and
imaginary parts of $f$.


*

*Why is $f^{(k)}\to f$ almost everywhere, not everywhere? Does this suggest that for BMO functions and in general harmonic analysis, we allow $f$ to have the value $\infty$?


*How does that make $|| f^{(k)} ||_{BMO} \to || f||_{BMO}$?
The book suggests that we need to use the fact that if $f$ and $g$ belong to $BMO$ then so do $min(f,g)$ and $max(f,g)$.


*Can you please briefly how I can generalize this to complex-valued case?
Thank you very much.
 A: I just manage to get one-sided estimate for 2).
First of all,

*

*Yes, since Lebesgue measurable requires defined a.e. and so if $f(x)=\infty$ on a set of measure zero, then the truncation valid a.e. only


*Note that with Lebesgue dominated convergence theorem, one has $\text{Avg}_{B}f^{(k)}\rightarrow\text{Avg}_{B}f$ and hence $|f^{(k)}(x)-\text{Avg}_{B}f^{(k)}|\rightarrow|f(x)-\text{Avg}_{B}f|$ a.e.
Fatou lemma gives
\begin{align*}
\dfrac{1}{|B|}\int_{B}|f(x)-\text{Avg}_{B}f|&\leq\liminf_{k\rightarrow\infty}\dfrac{1}{|B|}\int_{B}|f^{(k)}(x)-\text{Avg}_{B}f^{(k)}|\\
&\leq\liminf_{k\rightarrow\infty}\|f^{(k)}\|_{\text{BMO}},
\end{align*}
so $\|f\|_{\text{BMO}}\leq\liminf_{k\rightarrow\infty}\|f^{(k)}\|_{\text{BMO}}$.


*Note that with $f=\text{Re}f+i\text{Im}f$, then
\begin{align*}
\int_{B}|f(x)-\text{Avg}_{B}f|&=\int_{B}|\text{Re}f+\text{Im}f-\text{Avg}_{B}\text{Re}f-\text{Avg}_{B}\text{Im}f|\\
&\leq\int_{B}|\text{Re}f-\text{Avg}_{B}\text{Re}f|+\int_{B}|\text{Im}f-\text{Avg}_{B}\text{Im}f|,
\end{align*}
so it is still the matter of real-valued function taken into account.

A: 

*To cover the remaining part, fix $\epsilon>0$. Choose a ball $D$ such that
$$ ⨍_D\left|f-⨍_Df\right|\geq \sup_B ⨍_B\left|f-⨍_Bf\right|-\epsilon. $$
By dominated convergence (used twice),
$$ ⨍_D\left|f^{(k)}-⨍_Df^{(k)}\right|\to ⨍_D\left|f-⨍_Df\right|, $$
thus,
$$ \limsup_{k\to\infty}\|f^{(k)}\|_{BMO}\geq \limsup_{k\to\infty}\geq⨍_D\left|f^{(k)}-⨍_Df^{(k)}\right|\geq \|f\|_{BMO}-\epsilon\qquad\forall\epsilon>0, $$
hence,
$$ \limsup_{k\to\infty}\|f^{(k)}\|_{BMO}\geq \|f\|_{BMO}. $$
