# Convergence Theorems in the extended real numbers

I'm studying measure theory and usually functions take values in the extended real line. For most of the theorems, pointwise limits of such functions are considered.

Now I was wondering whether some limit theorems that hold for $$\mathbb{R}$$ also hold in the extended real numbers since the proofs of these I know most of the time rely on the absolute value metric in $$\mathbb{R}$$. Are there alternative proofs that can be generalized more easily?

In particular, I would like to know if the following theorems hold and what the best way to prove them is, ideally without studying a lot of topology since I am familiar with metric spaces but not with topology.

For sequences in $$\mathbb{R}$$:

(1) Let $$\lim \limits_{n \to \infty} a_{n}=a$$ and $$\lim \limits_{n \to \infty} b_{n}=b$$. Then $$a_{n} \leq b_{n}$$ $$\forall n \implies a \leq b$$

(2) Let $$\lim \limits_{n \to \infty} a_{n}=x$$ and $$\lim \limits_{n \to \infty} b_{n}=x$$. Then $$a_{n} \leq x_{n} \leq b_{n}$$ $$\forall n \implies \lim \limits_{n \to \infty} x_{n}=x$$

(3) Let $$x_{n}$$ be monotonically increasing and bounded above, then $$\lim \limits_{n \to \infty}= \sup x_{n}$$.

Any hints or references are much appreciated.

Thanks very much!

Edit: I've recently found a note on the extended real number system. It states that there exists a bijective, order preserving function $$f:[-\infty,+\infty] \to [-1,+1]$$ defined by $$\varphi(c)=\frac{c}{1+\lvert c \rvert}$$ for $$c \in \mathbb{R}$$, $$\varphi(-\infty)=-1$$ and $$\varphi(\infty)=1$$.

I'm pretty sure this is even a homeomorphism, so it preserves convergence. Am I correct that this proves that all the theorems above for $$\mathbb{R}$$ carry over to the extended real numbers?

• You know the proofs of them for $\mathbb{R}$? – withoutfeather Aug 24 '19 at 10:28
• @withoutfeather Yes, I have omitted them for brevity. – DerivativesGuy Aug 24 '19 at 10:31

Yes, they all hold on the extended real line $$\overline{\mathbb R}$$ and you do not need topology to prove them. Actually, you prove them basically as in $$\mathbb R$$. Consider the third statement, for instance. If $$\sup_nx_n\in\mathbb R$$, you prove that $$\lim_n=\sup_nx_n$$ as in $$\mathbb R$$. If each $$x_n$$ is equal to $$-\infty$$, then it is clear that both $$\lim_nx_n$$ and $$\sup_nx_n$$ are equal to $$-\infty$$. Otherwise, both $$\lim_nx_n$$ and $$\sup_nx_n$$ are equal to $$\infty$$

• Yes, I think I can see your argument. But one thing I'm having some problems with is that for limits in the extended real line I cannot use the usual definition of convergence in a metric space and have to use the more general topological definition (of course the two coincide in the real numbers. But to show that a sequence $x_{n}$ convergences then requires to me to show that the tail of the sequence is close to its limit x for any open set $U$ s.t. $x \in U$. – DerivativesGuy Aug 24 '19 at 10:44
• My proof is the same as stated on wikipedia, so I'm using the lub property (en.wikipedia.org/wiki/Monotone_convergence_theorem#Lemma_1). – DerivativesGuy Aug 24 '19 at 10:50
• You can define a topology on $\overline{\mathbb R}$, but you don't have to. For instance, you can say that a sequence $(x_n)_{n\in\mathbb N}$ converges to $\infty$ there if, for each $M\in\overline{\mathbb R}$, there is some $N\in\mathbb N$ such that $$(\forall n\in\mathbb N):n\geqslant N\implies x_n\geqslant M.$$And you can say that a sequence $(x_n)_{n\in\mathbb N}$ converges to $-\infty$ there if, for each $M\in\overline{\mathbb R}$, there is some $N\in\mathbb N$ such that $$(\forall n\in\mathbb N):n\geqslant N\implies x_n\leqslant M.$$ – José Carlos Santos Aug 24 '19 at 11:02
• Thanks for your help! I've seen this definition before, but it seemed to kind of artificially solve the issue, but then I've also read that this definition follows naturally from the topological definition. I might try to prove this, but I'm afraid I would have to study topology some more to do that. – DerivativesGuy Aug 24 '19 at 11:35

Give $$\overline{R}$$ the order topology, as I describe here.

Then in any ordered topological space $$(X,\le)$$, the set $$U=\{(x,y) \in X^2: x \le y \}$$ is closed in $$X^2$$ (in the product topology on $$X^2$$ of course), the proof for $$X=\overline{\Bbb R}$$ is not hard: if $$(x,y) \notin U$$, this means that $$x > y$$. Find $$z \in \overline{\Bbb R}$$ such that $$x > z > y$$ and note that $$O_x=\{u \in \Bbb R: u > z\}$$ is an open neighbourhood of $$x$$ in $$\overline{\Bbb R}$$, $$O_y=\{u \in \Bbb R: u < z\}$$ is an open neighbourhood of $$y$$ in $$\overline{\Bbb R}$$ and $$(O_x \times O_y) \cap U = \emptyset$$, showing that $$U$$ is closed in the product topology.

A closed set is sequentially closed, so $$\forall n: a_n \le b_n$$ means $$(a_n,b_n) \in U$$ for all $$n$$ and so if $$a_n \to a$$ and $$b_n \to b$$, $$(a_n,b_n) \to (a,b)$$ in $$X^2$$ and so $$(a,b) \in U$$ as well, so $$a \le b$$.

It follows completely from standard order topology facts.

• Thanks for your answer. I think I need to study some more topology to follow your proof though. – DerivativesGuy Aug 27 '19 at 19:30
• Could you elaborate on why showing that U has empty intersection with $O_{x} \times O_{y}$ proves that U is closed? I've had some time to study a bit of topology recently, but still can't understand this part of the proof. Thanks. – DerivativesGuy Sep 22 '19 at 18:50
• @DerivativesGuy A set $C$ is closed iff for all $x \notin C$ there is some open neighbourhood $O_x$ that is disjoint from $C$. It’s a well-known characterisation of closed sets. – Henno Brandsma Sep 22 '19 at 20:59
• Yes, this basically uses the fact that C is closed if its completment is open. So I think I need to show that $\forall x \in U$ $\exists$ a basis element $B$ s.t. $x \in B$ and $B \cap U = \emptyset$. But I don't understand how the argument above proves that C is closed since you do not consider any point not in U and any basis element. Sorry if I'm missing something here. – DerivativesGuy Sep 23 '19 at 16:34
• @DerivativesGuy I show that for any $(x,y) \notin U$ we have an open neighbourhood missing $U$ so $U$ is closed using the criterion I stated, with $U$ in the role of $C$. – Henno Brandsma Sep 23 '19 at 16:45