A generalization about the density of $\mathcal C_c(X, E)$ in $\mathcal L_p (X, \mu, E)$ when $E$ is a Banach space

Question

Let $X$ be a metric space, $\mu$ a $\sigma$-finite non-negative Borel measure on $X$, and $(E, |\cdot|)$ a Banach space. Let $\mathcal L_p := \mathcal L_p (X, \mu, E)$ and $\|\cdot\|_{\mathcal L_p}$ be its semi-norm. Here we use Bochner integral. Let $\mathcal C_c :=\mathcal C_c(X, E)$ be the space of all $E$-valued continuous functions on $X$ with compact supports. It is well-known that

Theorem If $X$ is locally compact separable, then $\mathcal C_c$ is dense in $\big (\mathcal L_p, \|\cdot\|_{\mathcal L_p} \big)$ for all $p \in [1, \infty)$.

I would like to ask if above result can be further strengthened, i.e.,

Let $X$ be locally compact separable. For each $f \in \mathcal L_p$ with $p \in [1, \infty)$ and $\varepsilon>0$, there is $g \in \mathcal C_c$ such that $$ \color{red}{|g| \le |f|} \quad \mu\text{-a.e.} \quad \quad \text{and} \quad \quad \| f-g \|_{\mathcal L_p} < \varepsilon. $$

This result, if true, generalizes this lemma which itself generalizes another lemma.

Thank you so much for your elaboration!

Answer 1

Below is a counter-example taken from this thread. It works even when $\mathcal C_c$ is replaced by $\mathcal C$, the space of all continuous functions from $X$ from $E$.

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Let $X:=[0, 1]$, $E:=\mathbb R$, and $\mu$ the Lebesgue measure on $[0, 1]$. Let $C$ be the fat Cantor set. Then $C$ is closed and $\mu(C) =\frac{1}{2}$. Let $f :=1_C$. Then $\| f \|_{\mathcal L_1} = \frac{1}{2}$.

Let $g \in \mathcal C$ such that $|g| \le |f|$ $\mu$-a.e. This implies there is a $\mu$-null subset $N$ of $X$ such that $|g| \le |f|$ on $X \setminus N$. Assume $g (a) \neq 0$ for some $a \in X$. By continuity of $g$, there is an open interval $I$ of $X$ such that $g\neq0$ on $I$. Then $f>0$ on $I \setminus N$. So $(I \setminus N) \subset C$ and thus $I \subset (C \cup N)$.

Notice that $I \setminus C$ is non-empty and open, so there is a non-empty open interval $J$ such that $J \subset I$ and $J \cap C = \emptyset$. As such, $J \subset N$ and thus $\lambda (N)>0$. This is a contradiction. As such, $g =0$ everywhere. Hence any $g \in \mathcal C$ can not approximate $f$ from below.