Density of $L^\infty(\Omega)h$ in $L^p(\Omega)$ where $h \in L^p(\Omega)$

Question

Let $(\Omega,\mu)$ be a finite measure space. Suppose $1\leq p <\infty$. Let $h$ be an element of $L^p(\Omega)$ with $h >0$ a.e..

How show that the subspace $L^\infty(\Omega)h=\{ f h\ :\ f\in L^\infty(\Omega)\}$ is dense in $L^p(\Omega)$?

Answer 1

Put $E_N = \{x \in \Omega: |h(x)|^p \geq 1/N\}$ for $N \in \mathbb{N}$. Observe that $E_N \subset E_{N+1}$, and $\Omega = \cup E_N$. For any $g \in L^p(\Omega)$, apply the Lebesgue's monotone convergence theorem to $\{|g|^p\chi_{E_N}\}$: $$ \lim_{N\rightarrow \infty} \int_{E_N}|g|^pd\mu = \|g\|_p^p. $$ Given $\epsilon > 0$, pick $N$ such that $\int_{\Omega - E_N}|g|^pd\mu < \epsilon/2$. The set $S$ of simple measurable functions in $L^p(E_N)$ is dense in $L^p(E_N)$, so there exists an $s \in S$ such that $$ \int_{E_N} |s-g|^pd\mu < \epsilon/2. $$ Set $s = 0$ outside $E_N$, then $\|s-g\|_p^p < \epsilon$. Since $1/|h|$ is bounded on $E_N$, $s \in L^\infty(\Omega)h$. Hence $L^\infty(\Omega)h$ is dense in $L^p(\Omega)$.

Note that we do not need $\mu(\Omega) < \infty$.

I guess another approach is to do the same thing to $g$ on $E_N$ as what we did to $h$ in the first part, so we can find a set $F\subset E_N$ such that $g$ is bounded on $F$, and $\int_{E_N-F}|g|^pd\mu < \epsilon/2$.