Simple functions with finite-measure support are dense in $L^p(\mu )$ for $1\le p<\mathrm{\infty }$

Choose a measurable representative $u$ of $f$ by [L3]. For each [L1, L3, given, choose, construct] $m\in {\mathbb{N}}_{\ge 1}$, [L1] gives a simple function $s_m$ with $\mathrm{\mid }{s}_m\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$ and $s_m(x)\to u(x)$ for every $x$. Define $t_m:=s_m{\mathbf{1}}_{\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}m\}}\mathrm{.}$ Then each $t_m$ is simple and $\mathrm{supp}(t_m)\subseteq \{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}m\}$.

Since $\mathrm{\mid }u{\mathrm{\mid }}^p$ is integrable, [step 1.1, algebra] $\mu (\{\mathrm{\mid }u\mathrm{\mid }\ge 1\mathrm{/}m\})\le m^p\int \mathrm{\mid }u{\mathrm{\mid }}^{p}d\mu <\mathrm{\infty },$ so every $t_m$ has finite-measure support. Also $t_m(x)\to u(x)$ for every $x$: if $u(x)=0$ then eventually both sides are $0$, and if $u(x)\mathrm{\ne }0$ then ${\mathbf{1}}_{\{\mathrm{\mid }u(x)\mathrm{\mid }\ge 1\mathrm{/}m\}}=1$ for all large $m$. Finally $\mathrm{\mid }u-t_m{\mathrm{\mid }}^p\le (2\mathrm{\mid }u\mathrm{\mid }{)}^p,$ whose right-hand side is integrable.

Applying [L2] to $\mathrm{\mid }u-t_m{\mathrm{\mid }}^p$ gives [L2, step 2.1] $\mathrm{\parallel }f-t_m{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }u-t_m{\mathrm{\mid }}^{p}d\mu ⟶0.$ So the simple functions with finite-measure support are dense in $L^p(\mu )$.