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:

$\norm {f - g}_p < \epsilon$

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Proof

For $n \in \N$, we define $D_n : \R_{\ge 0} \to \R_{\ge 0}$ by:

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

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.

By Measurable Function is Simple Function iff Finite Image Set:

$f_n \in \map \EE \Sigma$

By $(2)$, for all $x \in X$:

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

In view of $(1)$, we have for all $x \in X$:

$\size {\map {f_n} x} \le \size {\map f x}$

In particular, $f_n \in \map {\LL^p} \mu$.

Moreover, for all $x \in X$:

$\size {\map {f_n} x - \map f x} \le \size {\map f x} + \size {\map {f_n} x} \le 2 \size {\map f x}$

Thus, by Lebesgue's Dominated Convergence Theorem:

$\ds \lim_{n \mathop \to \infty} \norm {f_n - f}_p = \paren {\int \lim_{n \mathop \to \infty} \size {f_n - f}^p \rd \mu }^{1/p} = 0$

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $12.11$