Space of Simple P-Integrable Functions is Everywhere Dense in Lebesgue Space

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $p \in \R$, $p \ge 1$.

Let $\map {\LL^p} \mu$ be Lebesgue $p$-space for $\mu$.

Let $\map \EE \Sigma \cap \map {\LL^p} \mu$ be the space of $\Sigma$-simple, $p$-integrable functions.

Then $\map \EE \Sigma \cap \map {\LL^p} \mu$ is everywhere dense in $\map {\LL^p} \mu$ with respect to the seminorm topology generated by the singleton $\set {\norm {\; \cdot \;}_p }$.

That is, for all $f \in \map {\LL^p} \mu$ and $\epsilon > 0$ there exists a $g \in \map \EE \Sigma \cap \map {\LL^p} \mu$ such that:
 * $\norm {f - g}_p < \epsilon$

Proof
For $n \in \N$, we define $D_n : \R_{\ge 0} \to \R_{\ge 0}$ by:
 * $\map {\Delta _n} y := \begin{cases}

\dfrac k {2^n} & : y \in \hointr {\dfrac k {2^n} } {\dfrac {k + 1} {2^n} }, k = 0, 1, \ldots, 2^{2 n}-1 \\ 0 & : y \ge 2^n \end{cases}$

Clearly, for all $n \in \N$:
 * $(1): \quad \forall y \in \R_{\ge 0} : 0 \le \map {\Delta_n} y \le y$

We also have:
 * $(2): \quad \ds \forall y \in \R_{\ge 0} : \lim_{n \mathop \to \infty} \map {\Delta_n} y = y$

since:
 * $\forall y \in \hointr 0 {2^n} : 0 \le y - \map {\Delta_n} y \le \dfrac 1 {2^n}$

Let:
 * $\map {f_n} x := \map \sgn {\map f x} \map {\Delta_n} {\size {\map f x} }$

where:
 * $\map \sgn \cdot$ denotes the signum function
 * $\size {\, \cdot \,}$ denotes the absolute value.

In view of $(1)$ and $(2)$, we have for all $x \in X$:
 * $\size {\map {f_n} x} \le \size {\map f x}$

and:
 * $\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$

Moreover, by Measurable Function is Simple Function iff Finite Image Set:
 * $f_n \in \map \EE \Sigma$

In view of Riesz's Convergence Theorem, it suffices to show:
 * $\ds \lim_{n \mathop \to \infty} \norm {f_n}_p = \norm f_p$

This follows from Lebesgue's Dominated Convergence Theorem.