# A question about the proof of the simple approximation lemma

by Artusartos
Tags: approximation, lemma, proof, simple
 P: 248 The Simple Approximation Lemma Let f be a measurable real-valued function on E. Assume f is bounded on E, that is, there is an $M \geq 0$ for which $|f|\leq M$ on E. Then for each $\epsilon > 0$, there are simple functions $\phi_{\epsilon}$ and $\psi_{\epsilon}$ defined on E which have the following approximation properties: $\phi_{\epsilon} \leq f \leq \psi_{\epsilon}$ and $0 \leq \psi_{\epsilon} - \phi_{\epsilon} < \epsilon$ on E. Proof: Let (c,d) be an open, bouned interval that contains the image of E, f(x), and $c=y_0 < y_1 < ... < y_n = d$ be a partition of the closed, bouned interval [c,d] such that $y_{k}-y_{k-1} < \epsilon$ for $1 \leq k \leq n$. $$I_k = [y_{k-1}, y_k)$$ and $$E_k = f^{-1}(I_k)$$ for $1 \leq k \leq n$ Since each $I_k$ is an inteval and the function f is measurable, each set $E_k$ is measurable. Define the simple functions $\phi_{\epsilon}$ and $\psi_{\epsilon}$ on E by $$\phi_{\epsilon} = \sum^{n}_{k=1} y_{k-1} . \chi_{E_k}$$ and $$\psi_{\epsilon} = \sum^{n}_{k=1} y_{k} . \chi_{E_k}$$ Let x belong to E. Since $f(E) \subseteq (c,d)$, there is a unique k, $1 \leq k \leq n$, for which $y_{k-1} \leq f(x) < y_k$ and therefore $$\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)$$. But $y_k - y_{k-1} < \epsilon$, and therefore $\phi_{\epsilon}$ and $\psi_{\epsilon}$ have the required approximation properties. My Question: "Let x belong to E. Since $f(E) \subseteq (c,d)$, there is a unique k, $1 \leq k \leq n$, for which $y_{k-1} \leq f(x) < y_k$ and therefore $$\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)$$." So if we choose any x, we will aways be able to find $y_k$ and $y_{k-1}$ such that $$\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)$$, right? But the theorem tell us that we need to find $\phi_{\epsilon}$ and $\psi_{\epsilon}$ so that $\phi_{\epsilon}$ is less than all possible values of f(x) and that $\psi_{\epsilon}$ is greater than all possible values of f(x) (not specific for any x you choose, so you the proof gives you a different $y_k$ and $y_{k-1}$ for each x...but the theorem tell us that there is only one for the whole function. Also, how can the whole function be greater than $\phi_{\epsilon}$ and less than $\psi_{epsilon}$, if the difference is less than epsilon?)...so I don't understand this proof.