Homeomorphisms defined on extended real line, what are the properties preserved? I am doing some self study on homeomorphisms on my own, and hope to get some help on the following:
1) Is the function $f(x)= (2/\pi) arc tan(x)$ a homoemorphism from the extended reals to $[-1,1]$ ? Obviously, it is a homeomorphism from the reals to $(-1,1)$ but I do not know how best to deal with infinities?
2) If $g(x)= 1/|x|$ for $x \neq 0$ and $g(x)= \infty$ for $x=0$, then is $f(g(x))$ a homemorphism from the extended reals to [-1,1]? Is the composition of homeomorphisms also a homeomorphism when dealing with infinities? 
3) Is Borel measurability preserved through a homeomorphism? In particular the function f(g(x)) above, is it borel measurable in the extended reals?
Thanks in advance for any insight.
 A: For Q1, as you have noted, we know $f$ is a homeomorphism from $\mathbb{R}$ to $(-1, 1)$, so the problem is at the end points. I will leave you to check $f$ is a bijection, tell me if you want more details. So we are left to show both $f$ and $f^{-1}$ are continuous. One way to solve the problem is to observe both $[-\infty, +\infty]$ and $[-1, 1]$ has the order topology defined on it. Therefore, $[-\infty, +\infty]$ has $\{(a, b) \mid a, b \in \mathbb{R}\} \cup \{[-\infty, b) \mid b \in \mathbb{R}\} \cup \{(a, +\infty] \mid a \in \mathbb{R}\}$ as a base. Similarly, $[-1, 1]$ has $\{(a, b) \mid a, b \in [-1, 1]\} \cup \{[-1, b) \mid b \in [-1, 1]\} \cup \{(a, 1] \mid a \in [-1, 1]\}$ as a base.
Now take any basic open set $B$ in $[1, 1]$. If $B$ is of the form $(a, b)$, then $f^{-1}\Big((a, b)\Big) = (\tan \frac{\pi}{2}a, \tan \frac{\pi}{2}b)$, which is open. If $B$ is of the form $[-1, b)$, then $f^{-1}\Big([-1, b)\Big) = [-\infty, \tan \frac{\pi}{2}b)$, which is open. If $B$ is of the form $(a, 1]$, then $f^{-1}\Big((a, 1]\Big) = (\tan \frac{\pi}{2}a, +\infty]$, which is open. You can see the argument for $f^{-1}$ being continuous is analogous, so I will skip.
For Q2, to show $f \circ g: \mathbb{R} \to [-1, 1]$ is not a homeomorphism, it suffices to show $\mathbb{R}$ and $[-1, 1]$ are not homeomorphic (i.e. no homeomorphisms exists between $\mathbb{R}$ and $[-1, 1]$). Firstly, $\mathbb{R}$ is an unbounded subspace of $\mathbb{R}$, hence non-compact. Secondly, $[-1, 1]$ is a closed bounded subspace of $\mathbb{R}$, hence compact. Here we use the Heine–Borel theorem for $\mathbb{R}$. Since compactness is a topological invariant (i.e. two homeomorphic spaces must be either both compact or both non-compact), this shows $\mathbb{R}$ and $[-1, 1]$ are not homeomorphic.
For Q3, we have the following theorem.

Continuous function is Borel mesurable (see problem 46) Let $X, Y$ be topological spaces equipped with Borel $\sigma$-algebra and $f: X \to Y$ be a continuous function. Then the pre-image of any Borel set under $f$ is also Borel (i.e. $f$ is Borel measurable).

Claim Let $X, Y, Z$ be topological spaces equipped with Borel $\sigma$-algebra, $\beta: X \to Y$ be Borel measurable and $\gamma: Y \to Z$ be continuious. Then $\gamma \circ \beta$ is Borel mesurable.
Proof By the above theorem, $\gamma$ is Borel measurable, so we only need to show the composition of two Borel measurable functions is Borel measurable. Take any Borel set $B \subseteq Z$, we have $\gamma^{-1}(Z)$ is Borel since $\gamma$ is Borel measurable. In addition, $\beta^{-1}(\gamma^{-1}(Z))$ is Borel since $\beta$ is Borel measurable. Therefore, $(\gamma \circ \beta)^{-1}(Z) = \beta^{-1}(\gamma^{-1}(Z))$ is Borel. We conclude $\gamma \circ \beta$ is Borel mesurable.
Similarly, we have the composition of two continuous functions is continuous. In particular, the above works for homeomorphism as well since homeomorphism is continuous.
For your comment, yes $f \circ g$ is continuous. Since we have already shown $f$ is continuous, it suffices to show $g$ is continuous. Since $g: \mathbb{R} \to [-\infty, +\infty]$ is given by $x \mapsto \frac{1}{|x|}$, we have $g^{-1}$ maps $x \mapsto \pm \frac{1}{x}$. Take any basic open set $B$ from $[-\infty, +\infty]$. There are $7$ cases to consider. If $B$ is interval of type $(a, b)$ with $b \le 0$, then $g^{-1}\Big((a, b)\Big) = \emptyset$. If $B$ is interval of type $(a, b)$ with $b > 0, a > 0$, then $g^{-1}\Big((a, b)\Big) = (-\frac{1}{a}, -\frac{1}{b}) \cup (\frac{1}{b}, \frac{1}{a})$ which is open. If $B$ is interval of type $(a, b)$ with $b > 0, a \le 0$, then $g^{-1}\Big((a, b)\Big) = (-\infty, -\frac{1}{b}) \cup (\frac{1}{b}, +\infty)$ which is open. If $B$ is interval of type $(a, +\infty]$ with $a > 0$, then $g^{-1}\Big((a, \infty]\Big) = (-\frac{1}{a}, \frac{1}{a})$ which is open. If $B$ is interval of type $(a, +\infty]$ with $a \le 0$, then $g^{-1}\Big((a, \infty]\Big) = (-\infty, \infty)$ which is open. If $B$ is interval of type $[-\infty, b)$ with $b \le 0$, then $g^{-1}\Big([-\infty, b)\Big) = \emptyset$ which is open. If $B$ is interval of type $[-\infty, b)$ with $b > 0$, then $g^{-1}\Big([\infty, b)\Big) = (-\infty, -\frac{1}{b}) \cup (\frac{1}{b}, +\infty)$ which is open. So $g$ is continuous.
