Prove $L^{p_\theta }(X)\subset L^{p_0}(X)+L^{p_1}(X),$ where $\frac{1}{p_\theta }:=\frac{1-\theta }{p_0}+\frac{\theta }{p_1}.$ I am reading this page 
https://terrytao.wordpress.com/2009/03/30/245c-notes-1-interpolation-of-lp-spaces/
and getting stuck at the second inclusion (8), i.e.,
$$L^{p_\theta }(X)\subset L^{p_0}(X)+L^{p_1}(X),$$
where $\frac{1}{p_\theta }:=\frac{1-\theta }{p_0}+\frac{\theta }{p_1}, 0 \leq \theta \leq 1, 0<p_0<p_1\leq \infty.$ (All the assumptions are in the Lemma 9 on the link as above, actually $X$ can be just $\mathbb{R}$).
Could you please give me some references for the proof of this inclusion? I was trying to prove it but there is no progress.   
Thank you so much. 
 A: Loukas Grafakos, Classical Fourier Analysis (3rd Edition), page 14:
Fix $\gamma>0$, define $f_{\gamma}=f\chi_{|f|>\gamma}$ and $f^{\gamma}=f\chi_{|f|\leq\gamma}$, one can show that 
\begin{align*}
d_{f_{\gamma}}(\alpha)&=d_{f}(\alpha),~~~~\alpha>\gamma\\
&=d_{f}(\gamma),~~~~\alpha\leq\gamma\\
d_{f^{\gamma}}(\alpha)&=0,~~~~\alpha\geq\gamma\\
&=d_{f}(\alpha)-d_{f}(\gamma),~~~~\alpha<\gamma.
\end{align*}
Also that
\begin{align*}
\|f_{\gamma}\|_{L^{p}}^{p}&=p\int_{\gamma}^{\infty}\alpha^{p-1}d_{f}(\alpha)d\alpha+\gamma^{p}d_{f}(\gamma),\\
\|f^{\gamma}\|_{L^{p}}^{p}&=p\int_{0}^{\gamma}\alpha^{p-1}d_{f}(\alpha)d\alpha-\gamma^{p}d_{f}(\gamma),\\
\int_{\gamma<|f|\leq\delta}|f|^{p}d\mu&=p\int_{\gamma}^{\delta}d_{f}(\alpha)\alpha^{p-1}d\alpha-\delta^{p}d_{f}(\delta)+\gamma^{p}d_{f}(\gamma).
\end{align*}
If $f\in L^{p,\infty}(X,\mu)$, one can show that $f^{\gamma}\in L^{q}(X,\mu)$ for any $q>p$ and $f_{\gamma}\in L^{q}(X,\mu)$ for any $q<p$. Thus, $L^{p,\infty}\subseteq L^{p_{0}}+L^{p_{1}}$ when $0<p_{0}<p<p_{1}\leq\infty$.
Finally note that $L^{p_{0}}\subseteq L^{p_{0},\infty}$.
