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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!

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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 3 '12 at 2:27
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 3 '12 at 2:27
The result is in the context of extended reals. – azarel Feb 3 '12 at 2:33
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 3 '12 at 3:25
@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 4 '12 at 18:20
up vote 2 down vote accepted

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 $a\in\mathbb R$, $\{x:\sup_n f_n(x)>a\}=\{x:\exists n, f_n(x)>a\}=\cup_n\{x:f_n(x)>a\}$.

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