A question about the proof of the simple approximation lemma


by Artusartos
Tags: approximation, lemma, proof, simple
Artusartos
Artusartos is offline
#1
Feb20-13, 04:31 PM
P: 245
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 [itex]M \geq 0[/itex] for which [itex]|f|\leq M[/itex] on E. Then for each [itex]\epsilon > 0[/itex], there are simple functions [itex]\phi_{\epsilon}[/itex] and [itex]\psi_{\epsilon}[/itex] defined on E which have the following approximation properties:

[itex]\phi_{\epsilon} \leq f \leq \psi_{\epsilon}[/itex] and [itex]0 \leq \psi_{\epsilon} - \phi_{\epsilon} < \epsilon[/itex] on E.

Proof: Let (c,d) be an open, bouned interval that contains the image of E, f(x), and [itex]c=y_0 < y_1 < ... < y_n = d[/itex] be a partition of the closed, bouned interval [c,d] such that [itex]y_{k}-y_{k-1} < \epsilon[/itex] for [itex]1 \leq k \leq n[/itex].

[tex]I_k = [y_{k-1}, y_k)[/tex] and [tex]E_k = f^{-1}(I_k)[/tex] for [itex]1 \leq k \leq n[/itex]

Since each [itex]I_k[/itex] is an inteval and the function f is measurable, each set [itex]E_k[/itex] is measurable. Define the simple functions [itex]\phi_{\epsilon}[/itex] and [itex]\psi_{\epsilon}[/itex] on E by

[tex]\phi_{\epsilon} = \sum^{n}_{k=1} y_{k-1} . \chi_{E_k}[/tex]

and [tex]\psi_{\epsilon} = \sum^{n}_{k=1} y_{k} . \chi_{E_k}[/tex]

Let x belong to E. Since [itex]f(E) \subseteq (c,d)[/itex], there is a unique k, [itex]1 \leq k \leq n[/itex], for which [itex]y_{k-1} \leq f(x) < y_k[/itex] and therefore [tex]\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)[/tex].

But [itex]y_k - y_{k-1} < \epsilon[/itex], and therefore [itex]\phi_{\epsilon}[/itex] and [itex]\psi_{\epsilon}[/itex] have the required approximation properties.

My Question:

"Let x belong to E. Since [itex]f(E) \subseteq (c,d)[/itex], there is a unique k, [itex]1 \leq k \leq n[/itex], for which [itex]y_{k-1} \leq f(x) < y_k[/itex] and therefore [tex]\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)[/tex]." So if we choose any x, we will aways be able to find [itex]y_k[/itex] and [itex]y_{k-1}[/itex] such that [tex]\phi_{\epsilon} (x) = y_{k-1} \leq f(x) < y_k = \psi_{\epsilon} (x)[/tex], right? But the theorem tell us that we need to find [itex]\phi_{\epsilon}[/itex] and [itex]\psi_{\epsilon}[/itex] so that [itex]\phi_{\epsilon}[/itex] is less than all possible values of f(x) and that [itex]\psi_{\epsilon}[/itex] is greater than all possible values of f(x) (not specific for any x you choose, so you the proof gives you a different [itex]y_k[/itex] and [itex]y_{k-1}[/itex] 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 [itex]\phi_{\epsilon}[/itex] and less than [itex]\psi_{epsilon}[/itex], if the difference is less than epsilon?)...so I don't understand this proof.
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