Extended pointwise product with zero at infinite factors

Let $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$ be the ordered set of The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined. Define a new total operation $\star :\bar{\mathbb{R}}\times \bar{\mathbb{R}}\to \bar{\mathbb{R}}$ as follows. If $a=0$ or $b=0$, put $a\star b=0$. If both factors are finite, put $a\star b=ab$ using real multiplication. In the remaining cases both factors are nonzero and at least one is infinite; put

$a\star b:=\{\begin{array}{ll}+\mathrm{\infty },& \mathrm{sgn}(a)\mathrm{sgn}(b)=1,\\ -\mathrm{\infty },& \mathrm{sgn}(a)\mathrm{sgn}(b)=-1,\end{array}$

where $\mathrm{sgn}(a)=1$ for $a>0$ and $\mathrm{sgn}(a)=-1$ for $a<0$ in the extended order. The clauses exhaust the domain and agree when a zero factor is finite, so they specify a unique value for every pair. Thus $0\star (\pm \mathrm{\infty })=0$, $(+\mathrm{\infty })\star (-2)=-\mathrm{\infty }$, and $(-\mathrm{\infty })\star (-\mathrm{\infty })=+\mathrm{\infty }$.

For any set $X$ and maps $f,g:X\to \bar{\mathbb{R}}$, their extended pointwise product with zero at infinite factors is the map $f\star g:X\to \bar{\mathbb{R}}$ defined by $(f\star g)(x)=f(x)\star g(x)$.

This explicitly chosen operation does not redefine the partial extended-real arithmetic of The extended real line $\bar{\mathbb{R}}=\mathbb{R}\cup \{-\mathrm{\infty },+\mathrm{\infty }\}$, its order, and the arithmetic that is left undefined, where $0\cdot \mathrm{\infty }$ is left undefined. It is not a limit rule or a field multiplication. No measurability theorem is part of this definition: such a theorem must specify the measurable structures and check the finite, zero, and positive or negative infinite factor sets separately.