I have the next question: given $(\Omega,\mathcal{F},\mu)$ a p.m.s. and $1\leq p<r<q$ then $$ \overline{L^p(\mu)\cap L^q(\mu)} = L^r(\mu) $$ with respect to $||\cdot||_r$.
I already know that $L^p(\mu)\cap L^q(\mu)\subset L^r(\mu)$. And in other post where this question was asked they said to consider $f\in L^r$ and $f_n = f\cdot\mathbb{1}_{\{n^{-1}\leq |f| \leq n\}}$ and that $f_n\in L^p(\mu)\cap L^q(\mu)$, but i dont understand why this is true, why does $f_n\in L^p(\mu)\cap L^q(\mu)$.
I hope someboy can help me with this.
The fact that $f_n$ belongs to $L^p\cap L^q$ follows from the inequalities $$ \lvert f_n\rvert^p=\lvert f_n\rvert^r\lvert f_n\rvert^{p-r}\leqslant \lvert f\rvert^r\lvert f_n\rvert^{p-r}\leqslant \lvert f\rvert^rn^{r-p} $$ and $$ \lvert f_n\rvert^q=\lvert f_n\rvert^r\lvert f_n\rvert^{q-r}\leqslant \lvert f \rvert^r\lvert f_n\rvert^{q-r}\leqslant \lvert f \rvert^r n^{q-r}. $$