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If $\varphi_1$ and $\varphi_2$ are simple function in $M(X,\Sigma)$, then

$\psi=sup\{\varphi_1,\varphi_2\}$, $\omega=inf\{\varphi_1,\varphi_2\}$

are also simple functions in $M(X,\Sigma)$.

proof: Since $\varphi_1$ and $\varphi_2$ are simple function in $M(X,\Sigma)$ they can be represent in the standard form

$\varphi_1=\sum_{i=1}^na_i\chi_{E_i}$ and $\varphi_2=\sum_{j=1}^mb_j\chi_{F_j}$

where the $a_i$ are distinct, $E_i$ are disjoint nonempty subsets of $X$ and are such that $X=\bigcup_{i=1}^n E_i$ and $b_j$ are distinct, $F_j$ are disjoint nonempty subsets of $X$ and are such that $X=\bigcup_{j=1}^m F_j$. Let $G_i=E_j\cap F_k$, $1\leq j\leq n$ and $1\leq k\leq m$, $c_i=max\{a_j,b_k \}$, $1\leq j\leq n$ and $1\leq k\leq m$.

My question is if $\psi=\sum_{i=1}^m c_i\chi_{G_i}$ is $sup\{\varphi_1,\varphi_2\}$ and if I can define the inf in the same way.

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  • $\begingroup$ Every finite set contains its supremum and infimum. $\endgroup$ – Henricus V. Mar 16 '16 at 23:39
  • $\begingroup$ You seem to be right. I have double checked your answer. Similarly you should be able to define the inf with a little thought. $\endgroup$ – астон вілла олоф мэллбэрг Mar 17 '16 at 0:01
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Let $\psi(x)=sup\{\varphi_1(x),\varphi_2(x)\}$. Then, either $\psi (x)=\varphi_1(x)$ or $\psi (x)=\varphi_2(x)$, and as the range of each is finite, so is that of $\psi$.

Similarly for $\omega$.

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  • $\begingroup$ I think that $\psi=\varphi_1$ or $\psi=\varphi_2$ if $sup=\{\varphi_1,\varphi_2\}=max\{\varphi_1,\varphi_2\}$ but nothing guarantees this. $\endgroup$ – user292200 Mar 17 '16 at 2:25
  • $\begingroup$ maybe I have not understood the question but if you define $\psi(x)=sup\{\varphi_1(x),\varphi_2(x)\}$ then this is the same as $max\{\varphi_1(x),\varphi_2(x)\}$. $\endgroup$ – Matematleta Mar 17 '16 at 2:28

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