$L^{1} \cap L^{\infty}$ is dense is in $L^{p}$

Question

Why is $L^{1} \cap L^{\infty}$ dense is in $L^{p}$?

My work: i use the $$\Vert f \Vert_{p} \leq \Vert f \Vert^{1/p}_{1} \Vert f \Vert_{\infty}^{1-1/p}.$$

Answer 1

Simple functions are dense in $L^p$ (by exactly the same argument that they are dense in $L^1$). These simple functions can always be taken to be bounded. In a $\sigma$-finite measure space, they can also be taken to have finite measure support. A bounded, measurable function with finite measure support is easily seen to be in $L^1 \cap L^\infty$.

Reference: Royden and Fitzpatrick, Chapter 18.