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According to my textbook, supremums of measurable functions exist and are measurable. But what about the sequence of functions $f_n: [0, 1] \to \mathbb{R}$ given by $f_n = n$? I don't think this sequence has a supremum but I do think all those functions are measurable. How can I reconcile this difference?

Thank you!

badatmath
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    When you are learning measure theory, you usually allow your functions to take on the values $\pm\infty$, right? So that's a way to make sense of $\sup_n n$. – Dylan Moreland Feb 03 '12 at 02:27
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    I think you need to better understand what does "supremum of a measurable function" mean and then you can understand this statement better. – Patrick Da Silva Feb 03 '12 at 02:27
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    The result is in the context of extended reals. – azarel Feb 03 '12 at 02:33
  • Got it, thanks! – badatmath Feb 03 '12 at 02:58
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    The first sentence should say that the supremum of a countable set of measurable functions is measurable. It is not true for an arbitrary set of measurable functions. – Jonas Meyer Feb 03 '12 at 03:25
  • @JonasMeyer: why not? – badatmath Feb 04 '12 at 15:52
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    @badatmath: First of all, do you agree that your textbook only shows that it is true for countably many functions? If so, the statement in your question could be corrected. Now, let $f$ be an arbitrary positive nonmeasurable function on $\mathbb R$. For each $x\in\mathbb R$, let $f_x:\mathbb R\to\mathbb R$ be defined by $f_x(x)=f(x)$, $f_x(y)=0$ if $y\neq x$. Then each $f_x$ is measurable and $f(y)=\sup\limits_{x\in\mathbb R}f_x(y)$ for all $y$. – Jonas Meyer Feb 04 '12 at 18:20
  • @JonasMeyer: oOoOoOoOo, nice counterexample! Thanks :D yeah I guess it only showed it for countable limits because only countable unions are additive? – badatmath Feb 04 '12 at 18:23
  • @badatmath: I am not sure exactly what you mean in your last comment, so perhaps I'm just rephrasing it: a $\sigma$-algebra is closed under countable unions but (usually) not arbitrary unions. E.g., an arbitrary set is the union of the singleton sets it contains, and (what is the same thing) an arbitrary characteristic function (even of a nonmeasurable set) is a sup of characteristic functions of singleton sets. – Jonas Meyer Feb 04 '12 at 18:28
  • @JonasMeyer: I see, thanks <3 – badatmath Feb 04 '12 at 19:13
  • @JonasMeyer Maybe you could collect the comment in an answer. – Davide Giraudo Feb 06 '12 at 12:16
  • @badatmath: Out of curiosity, what textbook? – Jonas Meyer Feb 08 '12 at 04:02
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    @JonasMeyer Real Analysis by Carothers – badatmath Feb 08 '12 at 21:48

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More precisely, what your textbook ought to say is that the pointwise supremum of a countable set of measurable functions is measurable.

As was pointed out by Dyland Moreland and azarel in the comments, this result is in a context where measurable functions are allowed to have codomain $[-\infty,+\infty]$ rather than $\mathbb R$. You gave an example of a countable set of measurable functions whose pointwise supremum is the constant function $f(x)=+\infty$, and $\{x:f(x)>a\}=[0,1]$ for all $a\in\mathbb R$, showing that $f$ is measurable.

The reason you cannot generally allow suprema of arbitrary sets of measurable functions while staying measurable is because you cannot generally allow arbitrary unions of measurable sets while staying measurable. If $E$ is a nonmeasurable set, then $E$ is a union of measurable sets, $E=\bigcup\limits_{x\in E}\{x\}$. Similarly, $\chi_E$ is a nonmeasurable function, but it is a supremum of measurable functions, $\chi_E(x)=\sup\limits_{y\in E}\chi_{\{y\}}(x)$.

To see more explicitly where countable unions come in, note that for all $a\in\mathbb R$, $\{x:\sup_n f_n(x)>a\}=\{x:\exists n, f_n(x)>a\}=\cup_n\{x:f_n(x)>a\}$.

Jonas Meyer
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