Truncation in Lorentz spaces I am reading a paper, whose author state the following: if $f \in L^{(q,\infty)}(\mathbb{R}^N)$, then $f_\delta \in L^p(\mathbb{R}^N)$ for every $p \in [1,q)$, where $\delta > 0$ and
$$
f_\delta = f\; \boldsymbol{1}_{\{x\in X: f(x) \geq \delta\}}.
$$
But is this really true?
 A: I will use the following fact

Lemma. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space and $p\in[1,+\infty)$. Then for measurable non-negative function $f$ and measurable set $A\in M$ we have
  $$
\int\limits_A f(x)^p d\mu(x)=p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt
$$
  where $F_{f,A}(t)=\mu(\{x\in A:f(x)>t\})$

Proof. Using Fubini theorem for positive functions we can say that
$$
\int\limits_A f(x)^p d\mu(x)=
\int\limits_A \int\limits_{(0,f(x))}pt^{p-1}dt d\mu(x)=
\int\limits_A \int\limits_{(0,+\infty)}pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)dt d\mu(x)=
$$
$$
\int\limits_{(0,+\infty)}\int\limits_A pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
\int\limits_{(0,+\infty)}pt^{p-1}\int\limits_A \boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
$$
$$
\int\limits_{(0,+\infty)}pt^{p-1}\mu(\{x\in A:f(x)>t\}) dt =
\int\limits_{(0,+\infty)}pt^{p-1}F_{f,A}(t) dt
$$

Theorem. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space, and $f\in L^{(q,\infty)}(X,M,\mu)$. Then for all $\delta>0$ the function $f_\delta = f\; \boldsymbol{1}_{\{x\in X:f(x)>\delta\}}$ is in $L^p(X,M,\mu)$. 

Proof. From definition of quasi-norm in Lorentz space we see that there exist $C>0$ such that
$$
\mu(\{x\in X:|f(x)|>t\})\leq \frac{C}{t^q}
$$
Denote $A=\{x\in X: f(x)>\delta\}$. Since $\delta>0$, then the function $f_\delta$ is non-negative and measurable. Then from previous lemma for $p\in[1,q)$ we have
$$
\Vert f_\delta\Vert_p^p=
\int\limits_{X} |f_\delta(x)|^pd\mu(x)=
\int\limits_A f(x)^pd\mu(x)=
p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt
$$
Note that for $t\in(0,\delta)$ we have $\mu(\{x\in A:f(t)>t\})=0$ so
$$
\Vert f_\delta\Vert_p^p=
p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,A}(t)dt\leq
p\int\limits_{[\delta,+\infty)}t^{p-1}F_{f,X}(f)dt\leq
p\int\limits_{[\delta,+\infty)}t^{p-1}\frac{C}{t^q}dt=
Cp\int\limits_{[\delta,+\infty)}\frac{1}{t^{q+1-p}}dt
$$
Since $p<q$ the last integral is convergent, so $\Vert f_\delta\Vert_p <+\infty$ and $f_\delta\in L^p(X,M,\mu)$.
