This homework problem involves showing that if $f,g$ are measurable simple functions, then so is $f+g$ and $f\,g$ - without using:

1) $\{x \in A: (f+g)(x) < t\} = \bigcup_{r\in\mathbb{Q}} \left[ \{x\in A: f(x) < r\} \cap \{x \in A: g(x) < t-r\} \right]$, for $(X,\mathcal{A})$, $X \supseteq A \in \mathcal{A}$, $f,g$ are $[0,{}^+\infty]$-valued measurable functions on $A$.

2) using $f-g$ as $f+(-1)\,g$, or $f^2$ proofs whereby one uses the set $\{x\in A: f^2(x) < t \} = \{ x\in A: -\sqrt{t} < f(x) < \sqrt{t} \}$

(I hope this is clear!)

So I believe (think) that a simple function is: $f := \sum_{k=1}^n\,a_k\,\chi_{E_k}$, where $E_k = f^{-1}(\{a_k\}) \in \mathcal{A}$ (though I don't think it has to belong to the algebra).

I am supposed to show these claims without using the two enumerations above in two different ways, one using simple functions, and the other using:

3) $f \vee g := \max(f(x),g(x)) \leftrightarrow \{x \in A: (f \vee g)(x) \leq t\} = \{x \in A: f(x) \leq t\} \cap \{x \in A: g(x) \leq t \}$

4) $f \wedge g \cdots \{ \} = \{ \} \cup \{ \}$

5) This theorem claims: $(X,\mathcal{A})$ - measurable space, $X \supseteq A \in \mathcal{A}$, $\{f_n\}$ be a sequence of $[{}^-\infty,{}^+\infty]$-valued measurable functions on $A$. Then (a) functions $\sup_n f_n$ and $\inf_n f_n$ are measurable, (b) $\limsup_n f_n$ and $\liminf_n f_n$ are measurable, and (c) $\lim_n f_n$ is measurable, with a domain where $\limsup_n f_n = \liminf_n f_n$

6) The sets $A_{n,k} = \{x \in A: \frac{k-1}{2^n} \leq f(x) < \frac{k}{2^n} \}$, which is like a scaled (to $n$) `largest integer function' that is more and more refined with increasing $n$

These are utilized in the claim: $(X,\mathcal{A})$ - measurable space, $X \supseteq A \in \mathcal{A}$, and $f$ be a $[0,+\infty]$-valued measurable function on $A$. Then there is a sequence $\{f_n\}$ of $[0,+\infty]$-valued simple measurable functions on $A$ that satisfy (1) $f_1(x) \leq f_2(x) \leq \cdots$ and (2) $f(x) = \lim_n f_n(x)$, at each $x \in A$.

I am pretty clear on 3,4,5, and 6; as well as how 1 and 2 work and why the question doesn't want them used (it would be too easy to quote a theorem). Unfortunately, I just don't know how to start - in either direction.

For the simple way, I was thinking of something like $$ (f+g)(x) = \sum_{n=1}^k\,(a_k^f+a_k^g)\,\chi_{E_k} = \cdots, $$ $$ (f\,g)(x) = \sum_{n=1}^k\,(a_k^f\,a_k^g)\,\chi_{E_k} = \cdots .$$

As for the way utilizing 3-6 I have no clue!

This is question 5 out of Donald Cohn's book Measure Theory, Chapter 2 section 1 - page 57 - if interested.

I usually ask math.stackexchange because the answers are not given and I eventually get to understand the question - that is what I am hoping for now too!

Thanks much!

  • 5
    $\begingroup$ (i) Please don't YELL (all caps is yelling); (2) there is no need to "sign" your messages with your name, since your user name always appears in the bottom right. $\endgroup$ Oct 17, 2011 at 2:37
  • $\begingroup$ Two good points I will keep in mind :) $\endgroup$
    – nate
    Oct 17, 2011 at 2:52

1 Answer 1


I think that your approach is too complicated to prove this for simple functions.

Given $E_1,E_2,\ldots E_n$ a simple function $f$ has the form $$f=\sum_{k=1}^na_k\chi_{E_k},$$ where $a_1,\ldots,a_n$ are real numbers. I suggest the following exercises:

  1. Prove that there exist $b_1,b_2,\ldots ,b_m$, pairwise distinct, and $A_1,\ldots ,A_m$, pairwise disjoint, such that $$f=\sum_{k=1}^na_k\chi_{E_k}=\sum_{k=1}^mb_k\chi_{A_k}.$$
  2. Prove that if $f=a\chi_{E}$, $a\neq 0$, then $f$ is measurable if and only if $A$ is measurable (note that if $a=0$, $f$ is constant and then measurable).
  3. Prove that any simple function $f=\sum_{k=1}^na_k\chi_{E_k},$ is measurable if and only if $E_1,\ldots,E_n$ are measurable sets.
  4. Prove that sum of simple functions is a simple function, prove that the product of simple functions is a simple function.

To prove that the sum of measurable simple functions is simple, consider the sets that you get when finishing 4. Same thing for the product. Hope this helps.

  • $\begingroup$ @nate: feel free in ask me if something is unclear. $\endgroup$
    – leo
    Oct 17, 2011 at 5:54
  • $\begingroup$ <sub> @leo thanks - a lot of thought later and i think i got it! </sub> $\endgroup$
    – nate
    Oct 17, 2011 at 17:43
  • $\begingroup$ you are welcome @nate $\endgroup$
    – leo
    Oct 18, 2011 at 0:04

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