How to prove something using the extended reals? How can you have a rigorous proof using the extended real numbers? For example if a function, $f$, is bounded and continuous on $\mathbb{R}$ and another continuous function, $g$, is such that $\lim_{t\to\pm\infty}g(t) = 0$ can we treat $f(t)g(t)$ on $\mathbb{R}$ as $h(t):[-\infty,\infty]\to \mathbb{R}$
$$h(t)=  \bigg\{ \begin{array} &f(t)g(t) \text{ if }t\in(-\infty,\infty)\\
0 \text{ if }t=\pm\infty
\end{array}
$$
or does this require more steps to justify this claim? If we can work with this funciton, $h$,  as it is defined can we treat $[-\infty,\infty]$ as compact?

The space $[-\infty,+\infty]$ is compact if (as is usually done), one defines open sets by saying that every set of the form $(a,\infty]$ is a basic open neighborhood of $+\infty$, and similarly for $-\infty$.  That makes the piecewise defined function $h$ continuous on $[-\infty,+\infty]$.  Here I am relying on the fact that you said $f$ is bounded.  Without that, one would not necessarily have $\lim_{t\to\infty} h(t)=0$ (for example, if $f(t)=t$ and $g(t)=1/t$ for $t>{}$ some number, then $f(t)g(t)$ would not approach $0$ as $t\to\infty$).
Defining neighborhoods of $+\infty$ or $-\infty$ in the way specified above is essential equivalent to the definition of $\lim_{t\to\infty} h(t)$ usually given in calculus courses, which says $\lim_{t\to\infty} h(t) = H$ if for every $\varepsilon>0$, there exists $T$ such that for all $t>T$ we have $|h(t) - H|<\varepsilon$.
So then $h$ is a continuous function on a compact space.
