Let $G$ be a locally compact Hausdorff group with left modular character $\delta$. A discrete subgroup $\Gamma$ of $G$ is called a lattice if $\delta(x) = 1$ for all $x \in \Gamma$, and if the resulting left invariant measure on $G/\Gamma$ is finite.
If $G$ has a lattice, and I'm trying to understand why $G$ has to be unimodular. The proof comes down to showing that since $G/\Gamma$ has finite measure, so does $G/N$, where $N$ is the kernel of $\delta$. Here we have $\Gamma \subseteq N \subseteq G$, so this should have something to do with $G/N$ being a continuous image of $G/\Gamma$. Somehow we need to relate the measures on $G/N$ and on $G/\Gamma$.
Once we show that $G/N$ has finite measure, we use the result that if a topological group has finite Haar measure, then it is compact. Then $G/N$ surjects onto a compact subgroup of $(0,\infty)$, which must be trivial.
If $H$ is any locally compact Hausdorff group, the Riesz representation theorem gives a bijection between Radon measures on $H$ and positive linear functionals on $C_c(H)$, where a measure $\mu$ is sent to the linear functional $\Phi$ defined by
$$f \mapsto \int\limits_{H} f \space d\mu $$
It is a general result that $\mu$ is a left Haar measure if and only if $\Phi(L_x(f)) = \Phi(f)$ for all $f \in C_c(H)$, where $L_x(f)(g) = f(x^{-1}g)$.
Let $\pi: G/\Gamma \rightarrow G/N$ be the quotient map $g \Gamma \mapsto g N$. Let $\lambda$ be the $G$-invariant Radon measure on $G/\Gamma$. Define a positive linear functional $\Phi: C_c(G/N) \rightarrow \mathbb{C}$ by
$$\Phi(f) = \int\limits_{G/\Gamma}(f \circ \pi) \space d\lambda$$
To check that this is well defined, we just have to check that this integral converges. But this is obvious, because $f$ is bounded on $G/N$, hence $f \circ \pi$ is bounded on $G/\Gamma$, and $G/\Gamma$ has finite measure.
Let $\mu$ be the Radon measure on the locally compact Hausdorff space $G/N$ which corresponds to $\Phi$. Since $G/N$ is a topological group, and $\Phi(L_{xN}(f)) = \Phi(f)$ for all $f \in C_c(G/N)$ and $xN \in G/N$, we can conclude that $\mu$ is a left Haar measure on $G/N$. Then
$$\mu(G/N) = \int\limits_{G/N} \textrm{Char}(G/N) \space d\mu = \Phi(\textrm{Char}(G/N)) = \int\limits_{G/\Gamma} \textrm{Char}(G/\Gamma) \space d\lambda = \lambda(G/\Gamma) < \infty $$