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Claim: Let $X$ be a set and X be a $\sigma$-algebra. If $\phi_1$ and $\phi_2$ are simple functions, then the function $\psi$ defined by $$\psi=\max\{\phi_1,\phi_2\},$$ is also simple.

Proof. [Attempt] Let $$\phi_1=\sum^{m}_{k=1}a_k \chi_{E_k}$$ and $$\phi_2=\sum^{n}_{i=1}b_i \chi_{F_i}$$ be the standard representations of $\phi_1$ and $\phi_2$, respectively. Consider the collection $\{E_k\cap F_i|k\in\{1,2,...,m\}, i\in\{1,2,...,n\}\}$. Note that $\bigcup{E_k\cap F_i}=X$. On each $E_k\cap F_i$, $\psi$ takes on one exactly one value. Hence, $\psi$ takes on at most $nm$ values; therefore, $\psi$ is simple. $\blacksquare$

Did I miss anything here? Thank you!

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Why do you claim the union of the intersection is the whole space? The indicator function of the rationals is a simple function but your claim would not be true in this case. Also each your simple functions can take n+1 and m+1 values respectively when counting zero. –  Christian Bueno Jul 21 '13 at 3:00
    
Thank you for your response. It appears my understanding of simple functions is lacking. I will try to construct a proof using the facts Peter Tamaroff supplied below. –  dgc1240 Jul 21 '13 at 3:12
    
@ChristianBueno, I thought that union of intersections would indeed equal the whole space X for the following reason. (I will prove just one direction of the inclusion since the other is obvious.) Let $x\in X$. Then, $x\in E_k$ for exactly one $k$ since the collection $\{E_k\}$ partitions $X$. Similarly, $x\in F_i$ for exactly one $i$ since the collection $\{F_i\}$ partitions $X$. Hence, $x\in E_k\cap F_i$, which is a subset of $\bigcup E_k\cap F_i$. –  dgc1240 Jul 21 '13 at 3:27
    
@dgc1240 If I were you I would just answer my own question. –  Pedro Tamaroff Jul 21 '13 at 4:33
    
@dgc1240 If I define $\phi_1=\phi_2=\chi_\mathbb{Q}$ then $E_1=F_1=\mathbb{Q}$ are the only sets involved and $\bigcup E_k\cap F_i = \mathbb{Q} \neq \mathbb{R}$ –  Christian Bueno Jul 21 '13 at 4:40
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2 Answers

up vote 0 down vote accepted

$$\max\{\phi_1,\phi_2\}=\frac{1}{2}\left(\phi_1+\phi_2\right)+\frac{1}{2}\left|\phi_1-\phi_2\right|.$$

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Thank you! After proving that simple algebraic combinations of simple functions are simple, I simply (no pun intended) used the identity you supplied above. One final question: was my proof completely wrong? Since this is a self-study, I like to reflect on ways to improve. In particular, I like to know where incorrect proof went wrong. Thanks again! –  dgc1240 Jul 21 '13 at 22:06
    
@dgc1240 I think your proof is correct, but you should clarify what you take the union over. –  Michael Greinecker Jul 22 '13 at 7:25
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Notice that this involves exactly the same steps I detailed. For example, to prove that $|\phi_1-\phi_2|$ is simple, one has to show that $\{x:\phi_1(x)\geq\phi_2(x)\}$ s in the $\sigma$-algebra or something equivalent to it. –  Mariano Suárez-Alvarez Jul 22 '13 at 7:36
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  • Show that the sum and product of two simple functions is simple.
  • If $\phi$ and $\psi$ are simple functions, show that the set $U=\{x:\phi(x)\geq\phi(y)\}$ is a set in the $\sigma$ algebra.
  • Conclude what you want, using the fact that $$\max\{\phi,\psi\}=\phi\cdot\chi_U+\psi\cdot\chi_{X\setminus U}.$$
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