The Simple Function Approximation Lemma

Table of Contents

Lemma 1 (The Simple Function Approximation Lemma): Let

$f$

be a Lebesgue measurable function on a Lebesgue measurable set

$E$

. If

$f$

is bounded on

$E$

then for all

\epsilon > 0

there exists simple functions

\varphi_{\epsilon}

and

\psi_{\epsilon}

$E$

such that:
1)

\varphi_{\epsilon}(x) \leq f(x) \leq \psi_{\epsilon}(x)

for all

x \in E

.
2)

0 \leq \psi_{\epsilon} - \varphi_{\epsilon} < \epsilon

for all

x \in E

Proof: Let $$f$$ be a Lebesgue measurable function on a Lebesgue measurable set $$E$$ , and let $$f$$ be bounded on $$E$$ . Then there exists $m, M \in \mathbb{R}$ with $$m < M$$ such that for all $x \in E$ we have that:

(1)

\begin{align} \quad m \leq f(x) \leq M \end{align}

Consider the interval $$[m, M]$$ . Let $P = \{ m = x_0, x_1, ..., x_n = M \}$ be a partition of $$[m, M]$$ with $$m = x_0 < x_1 < ... < x_n = M$$ with the property that $\displaystyle{x_k - x_{k-1} < \epsilon}$ for each $k \in \{ 1, 2, ..., n \}$ and also, for each $$k$$ define:

(2)

\begin{align} \quad E_k = f^{-1} ([x_{k-1}, x_k)) \end{align}

Then each $$E_k$$ is a Lebesgue measurable set and the collection of sets $\{ E_1, E_2, ..., E_n \}$ are mutually disjoint. For each $k \in \{ 1, 2, ..., n \}$ define the functions $\varphi_k$ and $\phi_k$ on the sets $$E_k$$ by:

(3)

\begin{align} \quad \varphi_k(x) = x_{k-1} \quad \mathrm{and} \quad \psi_k(x) = x_k \end{align}

Then define the functions $\varphi_{\epsilon}$ and $\psi_{\epsilon}$ by:

(4)

\begin{align} \quad \varphi_{\epsilon}(x) = \left\{\begin{matrix} \varphi_1(x) = x_0 & \mathrm{if} \: x \in E_1 \\ \varphi_2(x) = x_1 & \mathrm{if} \: x \in E_2 \\ \vdots & \\ \varphi_n(x) = x_{n-1} & \mathrm{if} \: x \in E_n \end{matrix}\right. \end{align}

(5)

\begin{align} \quad \psi_{\epsilon}(x) = \left\{\begin{matrix} \psi_1(x) = x_1 & \mathrm{if} \: x \in E_1 \\ \psi_2(x) = x_2 & \mathrm{if} \: x \in E_2 \\ \vdots & \\ \psi_n(x) = x_n & \mathrm{if} \: x \in E_n \end{matrix}\right. \end{align}

Then $\varphi_{\epsilon}$ and $\psi_{\epsilon}$ are simple functions and for all $x \in E$ we have that $\varphi_{\epsilon}(x) \leq f(x) \leq \psi_{\epsilon}(x)$ , so (1) holds.

Furthermore, for each $k \in \{ 1, 2, ..., n \}$ we have that $\psi_{k}(x) - \varphi_k(x) = x_k - x_{k-1} < \epsilon$ for all $x \in E_k$ and so $0 \leq \psi_{\epsilon}(x) - \varphi_{\epsilon}(x) < \epsilon$ for all $x \in E$ . $\blacksquare$

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The Simple Function Approximation Lemma