# If $f,g$ are simple measurable functions, show that $f+g$ and $f\,g$ are too

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!

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(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. – Arturo Magidin Oct 17 '11 at 2:37
Two good points I will keep in mind :) – nate Oct 17 '11 at 2:52

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

Given $E_1,E_2,\ldots E_n$ a simple function $f$ have the form $$f=\sum_{k=1}^na_k\chi_{E_k},$$ where $a_1,\ldots,a_n$ are real numbers. I suggest you 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 finish 4. Same thing for the product. Hope this help you.

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@nate: feel free in ask me if something is unclear. – leo Oct 17 '11 at 5:54
<sub> @leo thanks - a lot of thought later and i think i got it! </sub> – nate Oct 17 '11 at 17:43
you are welcome @nate – leo Oct 18 '11 at 0:04