Function decomposition in harmonic analysis In the book "Weighted Norm Inequalities and Related Topics" by José García-Cuerva, J.-L. Rubio de Francia page 144 it was shown that for a measurable function $f$ and $t>0$
$$
|E_t|\leq \frac{C}{t}\int_{\{x\in\mathbb{R}^n:|f(x)|>t/2\}}|f(x)|dx,
$$
where $E_t=\{x\in\mathbb{R}^n:Mf(x)>t\}$, $M$ denotes the Hardy-Littlewood maximal operator.
In the proof they decompose $f$ such that $f=f_1+f_2$, where $f_1(x)=f(x)$ if $|f(x)|>t/2$, and $f_(x)=0$ otherwise. Then
$$
|E_t|\leq\{Mf_1(x)>t/2\}\leq\frac{3^n4^n}{t/2}\int_{\mathbb{R}^n}|f_1(x)|dx=\frac{C}{t}\int_{\{|f(x)|>t/2\}}|f(x)|dx.
$$
They use the weak $(1,1)$ boundedness of operator $M$. But don't be need to show that $f_1\in L^1$ ? 
As similar I read a proof about the boundedness of Riesz transform in Orlicz Spaces in the book "Weighted Inequalities in Lorentz and Orlicz Spaces" by Vakhtang Kokilashvili and Miroslav Krbec.
They also take $f$ from Orlicz space $L^{\Phi}$ and decompose $f$ as above and use the weak $(1,1)$ and strong $(p,p)$ boundedness of Riesz transform for $f_1$ and $f_2$, respectively. They also don't need to show that $f_1\in L^1$ and $f_2\in L^p$.
Are these cases trivial ? What is the point that I miss?
 A: For your first question, I think the point is that if $f_1\not\in L^1$, then
$$
\int |f_1|\,dx = \int_{\{|f(x)|>t/2\}} |f|\,dx = \infty,
$$
so the right hand side of the first inequality is $\infty$, and it is trivially satisfied.  
I can't say anything about the Orlicz space question; I'd have to see more details.  What I can say in some generality though is that for the "big" pieces in these decompositions, it's easier to show it's in $L^p$ for small $p$, and vice-versa for "smaller" pieces.  
EDIT: I decided to add a quick example to back up that last sentence.
Let $\mathcal{F}$ be the Fourier transform operator
$$
\mathcal{F}f(\xi) = \int_{\mathbb{R}^n} e^{ix\cdot \xi} f(x)\,dx.
$$
It is easy to show that $\mathcal{F}:L^1(\mathbb{R}^n)\to L^\infty(\mathbb{R}^n)$ is continuous, that is, 
$$
\|\mathcal{F}f\|_{L^\infty} \leq C\|f\|_{L^1}.
$$
Moreover, by Parseval's theorem we have that $\mathcal{F}:L^2(\mathbb{R}^n)\to L^2(\mathbb{R}^n)$ is also continuous because $\|\mathcal{F}f\|_{L^2} = \|f\|_{L^2}$.
We can use interpolation to show that $\mathcal{F}$ is continuous as an operator from $L^p$ to $L^{p'}$ where $p' = \frac{p}{p-1}$ and $1<p<2$.  Instead of giving a full proof I will give one of the ideas and a reference to a proof.  The idea is that if $f\in L^p$ for $1<p<2$, then we can write 
$$
f(x) = f(x)1_{\{|f(x)|>t\}} + f(x)1_{\{|f(x)|\leq t\}} = f_{>t}(x) + f_{\leq t}(x).
$$
Now $f_{>t}$ is the "big" piece, and it's in $L^1$.  We can check this by the layer-cake formula.  Indeed, 
$$
|\{|f(x)|>t\}| \leq t^{-p} \|f\|_{L^p}^p,
$$
so 
\begin{align*}
\|f_{>t}\|_{L^1} &= t |\{|f(x)|>t\}| + \int_t^\infty |\{|f_{>t}(x)|>\alpha\}|\,d\alpha \\
&\leq t^{1-p}\|f\|_{L^p}^p + \int_t^\infty \alpha^{-p} \|f\|_{L^p}^p\,d\alpha \\
&= \frac{p}{p-1} t^{1-p} \|f\|_{L^p}^p
\end{align*}
In http://home.wlu.edu/~baczkowskid/talks/Marcinkiewicz/marcinkiewicz.pdf, Lemma 2 there's a proof of the first equality.  Similarly, we can check that $f_{\leq t} \in L^2$:
\begin{align*}
\|f_{\leq t}\|_{L^2} &= 2\int_0^t \alpha |\{|f(x)|>\alpha\}|\,d\alpha \\
&\leq 2\int_0^t \alpha^{1-p} \|f\|_{L^p}^p\,d\alpha \\
&= \frac{2}{2-p} t^{2-p} \|f\|_{L^p}^p.
\end{align*}
So now we can use our above bounds on $\mathcal{F}$ to get estimates for $\|\mathcal{F}f_{>t}\|_{L^\infty}$ and $\|\mathcal{F}f_{\leq t}\|_{L^2}$. These can be combined to get a bound for $\|\mathcal{F}f\|_{L^{p'}}$, but this takes some more work.  This can be done in general for operators satisfying these weak-type bounds, and that's called the Marcinkiewicz interpolation theorem. (Though for the Fourier transform it's better to use the Riesz-Thorin theorem, which is a different story).
A final comment is that it's good to look at simple examples to see what kinds of functions are in $L^p$ for large $p$ but not small $p$ (answer: the ones that have mild peaks but bad decay) and functions that are in $L^p$ for small $p$ but not large $p$ (answer: the ones with bad peaks but rapid decay).  This helps to explain why chopping up a function into the large and small parts makes sense when you want to compare different $L^p$ norms.
