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Actually, I have ever posted about this theorem.

However, the reason why I asked this is I want to make sure about this theorem.


In my textbook referred, (I copied the theorem exactly same as the textbook because my poor English can change the original theorem),

$(4.13)$ Theorem

  1. Every function $f$ can be written as the limit of a sequence $\{f_k\}$ of simple functions.
  2. If $f\ge0$, the sequence can be chosen to increase to $f$, that is, chosen such that $f_k \le f_{k+1}$ for every $k$.
  3. If the function $f$ in either (i) or (ii) is measurable, then the $f_k$ can be chosen to be measurable.

However, my professor wrote on the board,

$f \geqslant 0$: mble $\Rightarrow$ $\exists$ simple function $f_k \nearrow f$.


I think $f$ need not be nonnegative. Should it be nonnegative like professor wrote?


(Add. Simple function's definition)

The characteristic function, $\chi_E(\mathbf{x})$, of a set $E\in\Bbb{R}^n$ is defined by $$\chi_E(\mathbf{x})=\cases{1\quad \text{if } \mathbf{x}\in E \\ 0\quad\text{if } \mathbf{x} \notin E}.$$

Clearly, $\chi_E$ is measurable if and only if $E$ is measurable. $\chi_E$ is an example of what is called a simple function: a simple function is one which assumes only a finite number of values, all of which are finite. If $f$ is a simple function taking (distinct) values $a_1, \cdots, a_N$ on (disjoint) sets $E_1, \cdots, E_N$, then $$f(\mathbf{x})=\sum^N_{k=1} a_k\chi_{E_k}(\mathbf{x})$$

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    $\begingroup$ The uparrow indicates that the sequence of functions is increasing, i.e. he uses the second part of the theorem. $\endgroup$ – Arthur Sinulis Jun 6 '16 at 6:20
  • $\begingroup$ @ArthurSinulis If $f$ can be negative, $f_k \nearrow f$ does not hold? I think although $f$ is not restricted as being nonnegative, we can find $\{ f_k \}$ satisfying $f_k \nearrow f$ and $f_k$'s are simple functions. $\endgroup$ – Danny_Kim Jun 6 '16 at 6:22
  • $\begingroup$ What's a "simple" function? $\endgroup$ – 5xum Jun 6 '16 at 6:35
  • $\begingroup$ @5xum I edited my post adding the definition of the simple functions. Sorry for my unskilled question. I thought simple function is used in general. $\endgroup$ – Danny_Kim Jun 6 '16 at 6:41
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If your definition of simple function does not allow infinite values, then you clearly can have a function with a single point mapped to $-\infty$ that is never bounded below by any simple function.

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  • $\begingroup$ Ah, I cannot even think about the negative infinity case. Thank you for giving me a good case. Can I ask one more question? If there is a condition, for every $k$, $f_k \gt -\infty$, then $f\ge0$ is not needed anymore, is it? $\endgroup$ – Danny_Kim Jun 6 '16 at 6:29
  • $\begingroup$ @Danny_Kim: I think it then depends on more details of your definition of simple function. The usual one on wikipedia means that even $x \mapsto x$ is not bounded below by any simple function. $\endgroup$ – user21820 Jun 6 '16 at 6:33
  • $\begingroup$ I added about the simple function's definition. In my textbook did not say $a_k$ is bounded, just said the number of terms are finite. $\endgroup$ – Danny_Kim Jun 6 '16 at 6:43
  • $\begingroup$ @Danny_Kim: Hmm something isn't right about your book's definition. What if $(a_0,a_1) = (\infty,-\infty)$? Since it doesn't explicitly exclude that possibility, I don't know what else it doesn't say. $\endgroup$ – user21820 Jun 6 '16 at 6:52

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