A question about the proof of the simple approximation lemma

[Artusartos]

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.

 

[fresh_42]

$f(x)$ as a whole lies between $\phi_\epsilon$ and $\psi_\epsilon$, which are step functions, a lower step function and an upper. For a single point $x_0$ we have only two single steps: the lower and the upper, and $f(x_0)$ lies between $y_{k-1}\chi_{E_k}$ and $y_{k}\chi_{E_k}$. However, all steps bands are close enough. Draw a picture.