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$\def\Rbf{\mathbf{R}}$ By the Heine-Borel theorem, the real line $\mathbf{R}=(-\infty,+\infty)$ with the standard topology is not compact.

By the Alexandroff extension, one has the one-point compactification $\Rbf\cup\{\infty\}$, which is homeomorphic to the circle $S^1$. On the other hand, the extended real line $[-\infty,\infty]$ with the order topology is compact; it is homeomorphic to the closed interval $[0,1]$.

When introducing the (abstract) Lebesgue integration theory, Rudin in his Real and Complex Analysis uses the extended real line instead of $\Rbf\cup\{\infty\}$ for the range of measurable functions.

Is it just a matter of convention or are there any deep reasons that one should use one not the other?

  • I hardly ever see $\mathbb R \cup {\infty}$ in measure theory. I see $[-\infty,\infty]$ and $[0,\infty]$ mostly. The last set has a useful and intuitive arithmetic. – zhw. Apr 30 '20 at 17:47

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Yes, it makes a big difference. The key is ordering.

Lebesgue integration theory relies critically on having a good ordering on the range space of your functions, and on the space where the integral takes its values. This can be seen in the very definition of the Lebesgue integral $\int f$ as the supremum of $\int g$ over all simple functions $g$ with $g \le f$. The word "supremum" involves the ordering on the space where the integral takes its values, and the condition $g \le f$ involves the ordering on the range space. You see it again in fundamental results like the monotone convergence theorem, Fatou's lemma, etc.

Now the ordering on $\mathbb{R}$ itself is pretty nice, what with the least upper bound property, but the extended reals are even nicer: in $[-\infty, \infty]$, every set has a least upper bound, every nondecreasing sequence converges, and so on. This means we can avoid a lot of special cases when talking about functions with singularities or whose integrals are infinite. The tradeoff is the algebra is less nice: $[-\infty, \infty]$ is no longer a field, and so we have to add some special cases for expressions like $0 \cdot \infty$ or $\infty + -\infty$. But that's manageable.

By contrast, $\mathbb{R}$ together with an "unsigned infinity" would be totally unsuitable, as it has no good ordering at all: you don't want to say either $0 < \infty$ or $0 > \infty$, so the trichotomy axiom fails. Then you have a problem with the definition of the integral: if you have a function $f$ that takes the value $\infty$ somewhere, you can't decide whether the simple function $g=0$ should be included in the supremum defining $\int f$.

Nate Eldredge
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  • Nice answer. I'm curious though, is there an integration theory that doesn't rely on orderings? – rubikscube09 Apr 30 '20 at 17:46
  • @rubikscube09: The Bochner integral comes to mind. Though even there, you're indirectly using the fact that the norm on your Banach space takes its values in a space $\mathbb{R}$ that has a nice ordering. I would venture to say that any theory that doesn't use the ordering of $\mathbb{R}$ at some level, isn't part of real analysis. – Nate Eldredge Apr 30 '20 at 17:54
  • Yes, that's a good example. Another example I would have thought of would be integration w.r.t projection valued measures in spectral calculus. But that, of course, comes from a duality argument, which goes back to $\mathbb{R}$. Moral of the story: $\mathbb{R}$ is the cornerstone of analysis (as one might expect). – rubikscube09 Apr 30 '20 at 17:56
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$ [-\infty, \infty] \text{ and } \mathbb R\cup\{\infty\} $ are two completely different animals if you consider the topology. Now a Borel $\sigma$-algebra on a topological space is the smallest sigma algebra containing the open sets of that topological space. So it has nothing to do with convention. I could have used any other set as the range of a measurable function. Recall the definition of a measurable function

Suppose $(X,\Sigma)$ and $(X',\Sigma')$ are two measurable spaces, and suppose that the $\sigma$-algebra $\Sigma'$ is generated by the family of sets $\Pi$. Then $f : X \rightarrow X'$ is $\Sigma/\Sigma'$ measurable if (and only if, trivially) $f^{-1}(E) \in \Sigma$ for all $E \in \Pi$.

And for topological spaces we take the sigma algebras to be Borel $\sigma$-algebra.

You can talk about a measurable function from any measurable space $(X,\Sigma)$ to a topological space $Y$ by taking the Borel $\sigma$ algebra on $Y$ denoted by $\mathcal{B}(Y)$.

$\mathbb R\cup\{\infty\}$ is obtained by "identifying" $+\infty$ and $-\infty$ of $ [-\infty, \infty] $ .(reference).

So the family of measurable functions from any measure space $(X,\Sigma)$ to $(\mathbb R\cup\{\infty\},\mathcal{B}(\mathbb R\cup\{\infty\}))$ and that to $ ([-\infty, \infty],\mathcal{B}( [-\infty, \infty]))$ are different and that is because of the difference in topology (as the $\sigma$-algebra depends on topology here).

So the difference you must have understood by now is that $ [-\infty, \infty] $ has a natural ordering which $\mathbb R\cup\{\infty\}$ doesn't. And that helps us in deriving the Lebesgue integration theory.

Noob mathematician
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    It's true that they are topologically different, but I don't think that's as much an issue as the difference in their order structure. See my answer. – Nate Eldredge Apr 30 '20 at 17:41
  • Indeed, $[-\infty, \infty]$ and $\mathbb{R} \cup {\infty}$ are Borel isomorphic, so as measurable spaces they are the same. There is no significant difference in the class of measurable functions you get, whether you use one or the other. The issues arise when you want to define the integral of those functions. – Nate Eldredge Apr 30 '20 at 17:44
  • Are they not different ? as measurable spaces? @NateEldredge – Noob mathematician Apr 30 '20 at 18:34
  • If I am going wrong somewhere please let me know, corrections are highly appreciated . – Noob mathematician Apr 30 '20 at 18:38
  • @Noobmathematician: The Borel sigma algebra contains $F_{\sigma}, G_{\delta}$ and many more such sets. So there is a possibility that differnt topologies give rise to the same Borel sigma algebra. – Bumblebee Apr 30 '20 at 18:44
  • @Noobmathematician: See this for a concrete example. This might pretty much be the case here as well. – Bumblebee Apr 30 '20 at 18:48
  • @Noobmathematician: They are the same in the sense that there is a bijection $f : \mathbb{R} \cup {\infty} \to [-\infty, \infty]$ such that $f, f^{-1}$ are measurable, i.e. $B$ is Borel in $[-\infty, \infty]$ iff $f^{-1}(B)$ is Borel in $\mathbb{R} \cup {\infty}$. This is well known though a little annoying to prove; there is probably a similar question elsewhere on this site but I can't find one right now. More generally, the same is true for any two uncountable Polish spaces. – Nate Eldredge Apr 30 '20 at 18:50
  • @Bumblebee don't you think $S^1$ and $[0,1]$ produces different Borel $\sigma$ algebra? – Noob mathematician Apr 30 '20 at 18:59