If $G$ is a locally compact group, there's a left invariant Borel measure on $G$, called Haar measure, which is unique up to multiplication by scalar. Denote it by $\mu$.
For any $g\in G$, $\mu_g(A)=\mu(Ag)$ is another left invariant measure, so $\Delta(g)=\frac{\mu(Ag)}{\mu(A)} \ge 0$ is a well-defined function, independent of $A$ as long as $\mu(A)>0$. It is called the modular function of $G$.
Writing $\frac{\mu(Agh)}{\mu(A)}=\frac{\mu(Agh)}{\mu(Ag)}\frac{\mu(Ag)}{\mu(A)}$, we see that $\Delta$ is multiplicative - as long as $\mu(Ag)\neq 0$, i.e. $\Delta(g)\neq 0$.
Why can't we have $\Delta(g)=0$?
Why is $\Delta$ continuous? I only know how to prove it in the case of Lie groups, where a formula $\Delta = \det Ad$ exists. I prefer a proof that doesn't use the construction of $\mu$, only its properties.
A Haar measure is positive on all nonempty open sets, and finite on all compact sets. If $A$ is a compact set with nonempty interior (such a set exists since $G$ is locally compact), we have $0 < \mu(A) < \infty$. But the right translation with $g$ is a homeomorphism of $G$, so $Ag$ is also a compact set with nonempty interior, whence $0 < \mu(Ag) < \infty$, and $\Delta (g) \neq 0$.
Further, a Haar measure is outer regular. Once again, let $A$ a compact set with nonempty interior. Let $\varepsilon > 0$ be given. By outer regularity, there is an open $U \supset A$ with $\mu(A) < \mu(U) < (1+\varepsilon)\mu(A)$. Since $A$ is compact, there is a neighbourhood $V$ of $e$ with $A\cdot V \subset U$. For $g\in V$ we then have
$$\Delta(g) = \frac{\mu(Ag)}{\mu(A)} \leqslant \frac{\mu(U)}{\mu(A)} < 1 + \varepsilon.$$
On the other hand, for $h \in V^{-1} = \{ v^{-1} : v\in V\}$, we have $Ah^{-1} \subset U$, or equivalently $A \subset Uh$, and hence
$$\Delta(h) = \frac{\mu(Uh)}{\mu(U)} \geqslant \frac{\mu(A)}{\mu(U)} \geqslant \frac{1}{1+\varepsilon} > 1 - \varepsilon.$$
So for $g \in W := V \cap V^{-1}$, we have $1-\varepsilon < \Delta(g) < 1+\varepsilon$. Since $W$ is a neighbourhood of $e$, that means $\Delta$ is continuous in $e$. The multiplicativity of $\Delta$ now yields the continuity on all of $G$.