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An interesting question came up in our analysis class today.

Let $S= \{f:\Bbb R \rightarrow \Bbb R \cup \{\infty\} \ | \ \{f>c\} \text{ is open for each } c \in \Bbb R\}.$ If $A$ is a nonempty indexing set and for each $a\in A$, let $f_a \in S$. Show that $\sup \{ f_a:a\in A \} \in S$.

Some classmates and I are confused as to what the supremum is doing here. If the supremum is just one $f_a$, for instance say $f_k$ for some $k \in A$ then it is trivial. It must mean something else. I think we are just tripped up on the meaning of the supremum of a sequence of functions.


Edit:

After clarification it must mean supremum regarding $x$.

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3 Answers 3

up vote 2 down vote accepted

The problem is that the supremum may not be reached, and even when it is it's not necessarily for the same index: consider $a=[0,1]$ and $f_a(x)=x-a$ for example.

Let $c\in\Bbb R$, we have to show that $\{x,\sup_{a\in A}f_a(x)>c\}$ is open. Let $x_0$ in this set. Two case:

  • $\sup_Af_a(x_0)$ is infinite, and in this case take $\{a_n\}\subset A$ such that $f_{a_n}(x_0)\to \infty$. There is $n$ such that $f_{a_n}(x_0)>c$, then we use the fact that $f_{a_n}\in S$ to get $r$ such that if $|x-x_0|<r$, then $\sup_{a\in A}f_a(x)>c$.
  • $\sup_Af_a(x_0)$ is finite; take $\{a_n\}\subset A$ such that $f_{a_n}(x_0)\to \sup_Af_a(x_0)$. For some $n$, as $\sup_{a\in A}f_a(x_0)$, we have $f_{a_n}(x_0)>c$ and we conclude in the same way.
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Ahhhh that logic does make sense. The way you put it confused me for a while but it's beginning to sink in. Thanks! –  KUSH Nov 20 '12 at 7:37
    
The only part that is now confusing me is when you say $f_{a_n} \in S$ leads to an $r$ s.t. if $|x-x_0|<r$ $\Rightarrow$ $\sup_{a\in A} f_a(x) > c$. How did you make that conclusion, or rather where did that $r$ come from? –  KUSH Nov 20 '12 at 7:42
1  
Since $f_{a_n}\in S$, $\{x,f_{a_n}>c\}$ is open and $x_0$ is in this set. –  Davide Giraudo Nov 20 '12 at 9:07
    
Got it, thanks for the response! –  KUSH Nov 20 '12 at 9:31

Let $A$ be any indexing set and for each $\alpha \in A$, $f_{\alpha}:\mathbb{R} \to \mathbb{R} \cup \{\infty\}$. The typical definition of $\sup_{\alpha \in A} f_{\alpha}$ is as a pointwise supremum; i.e., $(\sup_{\alpha \in A} f_{\alpha})(x) := \sup_{\alpha \in A} f_{\alpha}(x) $ where the righthand side is evaluating the supremum of the functions evaluated at $x$ (which is a collection of real numbers and hence it makes sense to take the supremum). This would be my guess as to what your instructor intended.

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Yes I think that must be the case. –  KUSH Nov 19 '12 at 22:37

It seems that your claim is not true for general $f$: Let $A$ be the lebesgue non-measurable set in $[0,1]$, and define, for each $\alpha\in A$, $$f_\alpha(x)=\begin{cases}1,&x=\alpha\\0,&x\neq\alpha\end{cases}$$ Then, each $f_\alpha(x)$ is measurable, while $$\sup_{\alpha\in A}f_\alpha(x)=\chi_A(x)$$ which is not measurable.

The set $S$ in your question is just the set of measurable functions on $R$.

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