Inclusion of $L^p$ and weak $L^p$ spaces Let $0<p_0<p_1<\infty$, $0<\theta<1$, and $1/p_\theta=(1-\theta)/p_0+\theta/p_1$. Show that
$$L^{p_\theta,\infty}(X)\subset L^{p_0}(X)+L^{p_1}(X).$$
Suppose that  $f\in L^{p_\theta,\infty}(X)$, we need to show that $f$ can be written as $f=f_0+f_1$ with $f_0\in L^{p_0}(X)$ and $f_1\in L^{p_1}(X)$. Without loss of generality, we can assume that $\|f\|_{L^{p_\theta,\infty}}=1$, then we have
$$\sup_{t>0}t\lambda_f(t)^{1/p_\theta}\leq 1,$$
where $\lambda_f(t):=\mu(\{x\in X:|f(x)|\geq t\}$ is the distribution function of $f$. I'm stuck here, I don't know how to determine $f_0$ and $f_1$?
 A: Let $0<p_0<p_1<\infty$, $\theta \in (0,1)$, $\frac{1}{p_\theta}=\frac{1-\theta}{p_0}+\frac{ \theta}{p_1}$ and $f\in L^{p_\theta,\infty}(X)$. For fixed $\delta>0$, set $$f_\delta(x) = f(x) \chi_{\left\{x\in X\ :\ |f(x)|>\delta \right\} }(x),$$ $$f^\delta(x) = f(x) \chi_{\left\{x\in X \ :\ |f(x)|\leq \delta \right\}} (x)$$ (this means that $f=f_\delta+f^\delta$). Since $f$ is measurable, the set $\left\{x\in X\ :\ |f(x)|>\delta \right\}$ is measurable, following the measurability of $\chi_{\left\{x\in X \ :\ |f(x)|> \delta \right\}}$ and consequently of $f_\delta$ (product of two measurable functions is measurable). Furthermore, $f^\delta$ is also measurable, because $f^\delta=f-f_\delta$.
I assert that $f_\delta \in L^{p_0}(X)$ and that $f^\delta \in L^{p_1}(X)$. In fact, due to the construction of these functions
\begin{align}
    &\lambda_{f_\delta}(t)=\begin{cases} \lambda_f(t)\quad if \quad t>\delta, \\ \lambda_f(\delta)\quad if \quad t\leq\delta , \end{cases}
    &\lambda_{f^\delta}(t)=\begin{cases} 0\quad if \quad t\geq \delta, \\ \lambda_f(t)-\lambda_f(\delta) \quad if \quad t <\delta. \end{cases}\nonumber\\
\end{align}
Thus, by Teonelli's Theorem and by the previous equation
\begin{align*}
\|f_\delta\|_{p_0}^{p_0}=\int_{X}|f_\delta(x)|^{p_0}\,d\mu(x)&=\int_{X}\left(p_0\int_{0}^{|f_\delta(x)|} t^{p_0-1} \,dt \right)\,d\mu(x) \\&=p_0\int_{0}^{\infty} \int_{\{x \in X\mid |f_\delta(x)|> t \}}t^{ p_0-1} \,d\mu(x)\,dt
\\&=p_0\int_{0}^{\infty} t^{p_0-1}\mu (\{x \in X\mid |f_\delta(x)|> t \}) ​​\,dt \\&= p_0\int_{0}^{\infty} t^{p_0-1}\lambda_{f_\delta}(t) \,dt\\&
=p_0\int_{0}^{\delta} t^{p_0-1}\lambda_{f}(\delta) \,dt + p_0\int_{\delta}^{\infty} t^{p_0-1 }\lambda_{f}(t) \,dt\\&
=\delta^{p_0}\lambda_{f}(\delta)+ p_0\int_{\delta}^{\infty} t^{p_0-1}\lambda_{f}(t) \,dt
\end{align*}
Furthermore, as $\sup_{t>0}t\lambda_f(t)^{1/p_\theta}=C<\infty$, we have $\lambda_f(t)\leq C^{p_\theta}t^ {-p_\theta}$ for any positive $t$ and so
\begin{align*}
\|f_\delta\|_{p_0}^{p_0}&\leq \delta^{p_0}(C^{p_\theta}\delta^{-p_\theta}) +p_0\int_{\delta} ^{\infty} t^{p_0-1} (C^{p_\theta}t^{-p_\theta}) \,dt\\& \leq C^{p_\theta}\delta^{p_0- p_\theta} + p_0 \cdot C^{p_\theta} \int_{\delta}^{\infty} t^{p_0-p_\theta-1}\,dt<\infty,
\end{align*}
because
\begin{align*}
{p_0-p_\theta-1}=p_0-\left(\frac{p_0p_1}{(1-\theta)p_0+\theta p_1}\right)-1<-1.
\end{align*}
Analogously, we have
\begin{align*}
\|f^\delta\|_{p_1}^{p_1}=\int_{X}|f^\delta(x)|^{p_1}\,d\mu(x)&=\int_{X}\left(p_1\int_{0}^{|f^\delta(x)|} t^{p_1-1} \,dt \right)\,d\mu(x)\\&=p_1\int_{0}^{\infty}\int_{\{x \in X\mid |f^\delta(x)|> t \}}t^{p_1-1} \,d\mu(x)\,dt\\&=p_1\int_{0}^{\infty} t^{p_1-1}\mu (\{x \in X\mid |f^\delta(x)|> t \}) ​​\,dt \\&= p_1\int_{0}^{\infty} t^{p_1-1}\lambda_{f^\delta}(t) \,dt\\&
=p_1\left(\int_{0}^{\delta} t^{p_1-1} (\lambda_f(t)-\lambda_f(\delta)) \,dt + \int_{\delta}^{\infty } t^{p_1-1}\cdot 0 \,dt\right)
\\&\leq -\delta^{p_1}\lambda_{f}(\delta) +p_1\int_{0}^{\delta} t^{p_1-1} (C^{p_\theta}t^ {-p_\theta}) \,dt\\& \leq p_1 \cdot C^{p_\theta} \int_{0}^{\delta} t^{p_1-p_\theta-1}\,dt< \infty,
\end{align*}
because
\begin{align*}
{p_1-p_\theta-1}=p_1-\left(\frac{p_0p_1}{(1-\theta)p_0+\theta p_1}\right)-1>-1.
\end{align*}
