$\overline{L^{n+1}}=L^p$ for $1\leq n\leq p \leq n+1<\infty$ for $\mu(X)<\infty$

Question

After reading this post, I began thinking that this is true. My thought process the following. Let $n\leq p \leq n+1$ for $n\geq 1$. $$\overline{L^{n+1}}=\overline{L^{n+1}\cap L^{n+1}}\subseteq \overline{L^n\cap L^{n+1}}=L^p.$$

Does this chain work?