[proofplan] We fix a left Haar measure $\mu$ and define $\Delta(g)$ as the scalar by which the pushforward of $\mu$ under right translation by $g$ differs from $\mu$. The existence and uniqueness of this scalar follow from the standard uniqueness of left Haar measure up to positive scalar. The homomorphism identity comes from comparing the pushforward by $R_{gh}$ with the successive pushforwards by $R_g$ and $R_h$, continuity follows by testing against one compactly supported [continuous function](/page/Continuous%20Function) of nonzero integral, and the final equivalence is exactly the assertion that all right translations preserve the chosen left Haar measure. [/proofplan] [step:Define the modular scaling factor using uniqueness of left Haar measure] Let $R_g:G\to G$ denote the smooth right translation map \begin{align*} R_g:G&\to G \end{align*} \begin{align*} x&\mapsto xg. \end{align*} For each $g\in G$, define the pushforward Radon measure $(R_g)_*\mu$ on $G$ by \begin{align*} (R_g)_*\mu(E)=\mu(R_g^{-1}(E)) \end{align*} for every Borel set $E\subset G$. We first check that $(R_g)_*\mu$ is left invariant. For $a\in G$, let $L_a:G\to G$ be the left translation map $L_a(x)=ax$. Since $L_a\circ R_g=R_g\circ L_a$, for every Borel set $E\subset G$ we have \begin{align*} (L_a)_*(R_g)_*\mu(E)=(R_g)_*\mu(L_a^{-1}(E)). \end{align*} By the definition of pushforward measure, this is \begin{align*} \mu(R_g^{-1}(L_a^{-1}(E))). \end{align*} Since $R_g^{-1}\circ L_a^{-1}=L_a^{-1}\circ R_g^{-1}$ and $\mu$ is left invariant, we obtain \begin{align*} \mu(R_g^{-1}(L_a^{-1}(E)))=\mu(L_a^{-1}(R_g^{-1}(E)))=\mu(R_g^{-1}(E)). \end{align*} Thus \begin{align*} (L_a)_*(R_g)_*\mu(E)=(R_g)_*\mu(E). \end{align*} So $(R_g)_*\mu$ is a left Haar measure on $G$. By the standard uniqueness theorem for left Haar measure on a locally compact group, any two left Haar measures on $G$ differ by multiplication by a unique positive scalar. Hence there exists a unique number $\Delta(g)\in\mathbb R_+$ such that \begin{align*} (R_g)_*\mu=\Delta(g)\mu. \end{align*} This defines a function $\Delta:G\to\mathbb R_+$. [/step] [step:Compare successive right translations to prove multiplicativity] Let $g,h\in G$. Since \begin{align*} R_h\circ R_g=R_{gh}, \end{align*} functoriality of pushforward measures gives \begin{align*} (R_{gh})_*\mu=(R_h)_*((R_g)_*\mu). \end{align*} Using the defining relation for $\Delta(g)$, we get \begin{align*} (R_h)_*((R_g)_*\mu)=(R_h)_*(\Delta(g)\mu). \end{align*} Pushforward is linear on positive scalar multiples of measures, so \begin{align*} (R_h)_*(\Delta(g)\mu)=\Delta(g)(R_h)_*\mu. \end{align*} Using the defining relation for $\Delta(h)$, this becomes \begin{align*} \Delta(g)(R_h)_*\mu=\Delta(g)\Delta(h)\mu. \end{align*} Therefore \begin{align*} (R_{gh})_*\mu=\Delta(g)\Delta(h)\mu. \end{align*} On the other hand, by definition of $\Delta(gh)$, \begin{align*} (R_{gh})_*\mu=\Delta(gh)\mu. \end{align*} The scalar relating a left Haar measure to $\mu$ is unique, so \begin{align*} \Delta(gh)=\Delta(g)\Delta(h). \end{align*} Thus $\Delta:G\to\mathbb R_+$ is a [group homomorphism](/page/Group%20Homomorphism). In particular, since $R_e=\operatorname{id}_G$, we also have $\Delta(e)=1$. [/step] [step:Test against a compactly supported function to prove continuity] Choose a function $f\in C_c(G;[0,\infty))$ such that \begin{align*} \int_G f(x)\,d\mu(x)>0. \end{align*} Such a function exists because $G$ is a Lie group, hence locally compact, and a left Haar measure is nonzero and finite on compact sets while assigning positive measure to nonempty open sets. Define the function $I:G\to\mathbb R$ by \begin{align*} I(g)=\int_G f(xg)\,d\mu(x). \end{align*} By the defining property of pushforward measure, \begin{align*} I(g)=\int_G f(y)\,d((R_g)_*\mu)(y). \end{align*} Since $(R_g)_*\mu=\Delta(g)\mu$, this gives \begin{align*} I(g)=\Delta(g)\int_G f(y)\,d\mu(y). \end{align*} Therefore, for every $g\in G$, \begin{align*} \Delta(g)=\frac{I(g)}{\int_G f(y)\,d\mu(y)}. \end{align*} It remains to note that $I$ is continuous. This is the standard continuity of the right translation action on compactly supported continuous functions: if $g_i\to g$ in $G$, then the functions $x\mapsto f(xg_i)$ converge uniformly to $x\mapsto f(xg)$ on a common compact set containing their supports for all sufficiently large $i$. Since $\mu$ is finite on compact subsets of $G$, integration against $\mu$ preserves this convergence. Hence \begin{align*} \lim_{i} I(g_i)=I(g). \end{align*} Thus $I$ is continuous, and since the denominator $\int_G f\,d\mu$ is a positive constant, $\Delta$ is continuous. [guided] We want to prove continuity of $\Delta$ as a function on $G$. The useful point is that $\Delta(g)$ is not just an abstract scalar: it can be recovered by integrating one fixed compactly supported function after translating it. Choose $f\in C_c(G;[0,\infty))$ with \begin{align*} \int_G f(x)\,d\mu(x)>0. \end{align*} This is possible because a Lie group is locally compact, so compactly supported continuous functions exist locally, and a nonzero left Haar measure gives positive measure to nonempty open sets. Define \begin{align*} I:G&\to\mathbb R \end{align*} \begin{align*} g&\mapsto \int_G f(xg)\,d\mu(x). \end{align*} The point of this definition is that the integrand $f(xg)$ is exactly the pullback of $f$ by the right translation $R_g$. By the defining property of pushforward measures, \begin{align*} \int_G f(xg)\,d\mu(x)=\int_G f(y)\,d((R_g)_*\mu)(y). \end{align*} The definition of the modular function gives \begin{align*} (R_g)_*\mu=\Delta(g)\mu. \end{align*} Substituting this into the previous identity yields \begin{align*} I(g)=\int_G f(y)\,d(\Delta(g)\mu)(y). \end{align*} Since $\Delta(g)$ is a positive scalar, it factors out of the integral: \begin{align*} I(g)=\Delta(g)\int_G f(y)\,d\mu(y). \end{align*} The denominator is nonzero by the choice of $f$, so \begin{align*} \Delta(g)=\frac{I(g)}{\int_G f(y)\,d\mu(y)}. \end{align*} Thus continuity of $\Delta$ reduces to continuity of $I$. Let $(g_i)$ be a net in $G$ converging to $g\in G$. The right translation action on compactly supported continuous functions is continuous in the following standard sense: the functions $x\mapsto f(xg_i)$ converge uniformly to $x\mapsto f(xg)$ on a compact set containing their supports for all sufficiently large $i$. Because $\mu$ is a Radon measure, it is finite on compact subsets of $G$. Therefore the [uniform convergence](/page/Uniform%20Convergence) on that compact set implies \begin{align*} \lim_i \int_G f(xg_i)\,d\mu(x)=\int_G f(xg)\,d\mu(x). \end{align*} Equivalently, \begin{align*} \lim_i I(g_i)=I(g). \end{align*} So $I$ is continuous. Since $\Delta$ is obtained from $I$ by division by the fixed positive number $\int_G f\,d\mu$, the function $\Delta:G\to\mathbb R_+$ is continuous. [/guided] [/step] [step:Identify unimodularity with right invariance] Suppose first that $\Delta(g)=1$ for every $g\in G$. Then for every $g\in G$, \begin{align*} (R_g)_*\mu=\Delta(g)\mu=\mu. \end{align*} Thus $\mu$ is right invariant. Since $\mu$ was already left invariant, $\mu$ is a bi-invariant Haar measure on $G$. Conversely, suppose that $G$ admits a Haar measure $\nu$ that is both left invariant and right invariant. Since $\nu$ is left Haar and $\mu$ is left Haar, uniqueness of left Haar measure gives a scalar $c\in\mathbb R_+$ such that \begin{align*} \nu=c\mu. \end{align*} For each $g\in G$, right invariance of $\nu$ gives \begin{align*} (R_g)_*\nu=\nu. \end{align*} Substituting $\nu=c\mu$ gives \begin{align*} (R_g)_*(c\mu)=c\mu. \end{align*} Since pushforward commutes with multiplication of a measure by a positive scalar, \begin{align*} c(R_g)_*\mu=c\mu. \end{align*} Dividing by $c>0$, we obtain \begin{align*} (R_g)_*\mu=\mu. \end{align*} By the defining uniqueness of the scalar $\Delta(g)$ in \begin{align*} (R_g)_*\mu=\Delta(g)\mu, \end{align*} we conclude that $\Delta(g)=1$. Since $g\in G$ was arbitrary, $\Delta$ is identically equal to $1$ on $G$. This proves that $G$ has a bi-invariant Haar measure if and only if $\Delta(g)=1$ for every $g\in G$. [/step]