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.

 

[math771]

Can you please explain it to me?

My

The key to understanding this proof lies in the definition of a simple function. A simple function is a function that takes on only a finite number of values. In this case, we are constructing \phi_{\epsilon} and \psi_{\epsilon} to be simple functions that approximate f on the set E. This means that for any given x \in E, there will be a unique k such that y_{k-1} \leq f(x) < y_k, and \phi_{\epsilon} (x) = y_{k-1} and \psi_{\epsilon} (x) = y_k.

As for your confusion about \phi_{\epsilon} and \psi_{\epsilon} being less than and greater than all possible values of f(x), this is simply a way of saying that they are upper and lower bounds for f(x) on the set E. The difference between \phi_{\epsilon} and \psi_{\epsilon} is less than \epsilon, which means that they are very close to each other and therefore provide a good approximation for f(x).

I hope this clarifies the proof for you. The key is to understand the definition of a simple function and how it is being used to approximate f on the set E.