The Simple Function Approximation Lemma

The Simple Function Approximation Lemma

 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 $\psi_{\epsilon}$ on $E$ 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)
• 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 $x \in E$. $\blacksquare$