[proofplan]
For each fixed element $g\in G$, we translate the measure $\nu$ on the right by defining a new measure $\nu_g(A)=\nu(Ag)$ on Borel subsets of $G$. Because right translation is a homeomorphism and left translations commute with this construction by associativity, $\nu_g$ is again a left Haar measure. The uniqueness theorem for left Haar measures up to positive scalar then gives $\nu_g=c(g)\nu$. Evaluating this identity on the whole compact group $G$ and using the normalization $\nu(G)=1$ forces $c(g)=1$, which is precisely right invariance.
[/proofplan]
[step:Define the right translate of the Haar measure]
Fix $g\in G$. Let $\mathcal B(G)$ denote the Borel $\sigma$-algebra of the underlying [topological space](/page/Topological%20Space) of $G$. Define
\begin{align*}
\nu_g:\mathcal B(G)&\to[0,\infty]
\end{align*}
\begin{align*}
A&\mapsto \nu(Ag).
\end{align*}
For this definition, note that the right translation map
\begin{align*}
R_g:G&\to G
\end{align*}
\begin{align*}
x&\mapsto xg
\end{align*}
is a smooth diffeomorphism, with inverse $R_{g^{-1}}$. Hence $R_g$ is a Borel automorphism, so $Ag=R_g(A)$ is Borel whenever $A\in\mathcal B(G)$. Since $R_g$ is bijective, it preserves disjointness of families of subsets, and countable additivity of $\nu_g$ follows from countable additivity of $\nu$. Thus $\nu_g$ is a Borel measure on $G$.
[/step]
[step:Verify that the translated measure is left Haar]
We prove that $\nu_g$ is left invariant. Let $h\in G$ and let $A\in\mathcal B(G)$. By associativity of the group operation,
\begin{align*}
(hA)g=h(Ag).
\end{align*}
Since $\nu$ is left invariant,
\begin{align*}
\nu_g(hA)=\nu((hA)g)=\nu(h(Ag))=\nu(Ag)=\nu_g(A).
\end{align*}
Moreover, $\nu_g$ is regular because it is obtained from the regular Borel measure $\nu$ by the Borel homeomorphism $R_g$, and it is nonzero because
\begin{align*}
\nu_g(G)=\nu(Gg)=\nu(G)=1.
\end{align*}
Thus $\nu_g$ is a left Haar measure on $G$.
[guided]
The purpose of introducing $\nu_g$ is to measure a set after moving it by right multiplication by $g$. To show that this new measure is a left Haar measure, we must verify that left translation does not change it.
Fix $h\in G$ and $A\in\mathcal B(G)$. The set $hA$ is the image of $A$ under the left translation map $L_h:G\to G$, $x\mapsto hx$, so $hA$ is Borel. By associativity, multiplying first on the left by $h$ and then on the right by $g$ gives the same set as multiplying $Ag$ on the left by $h$:
\begin{align*}
(hA)g=h(Ag).
\end{align*}
Therefore, using the definition of $\nu_g$ and then the left invariance of $\nu$,
\begin{align*}
\nu_g(hA)=\nu((hA)g)=\nu(h(Ag))=\nu(Ag)=\nu_g(A).
\end{align*}
This proves left invariance of $\nu_g$.
We also need the remaining Haar-measure properties. Since $R_g:G\to G$ is a homeomorphism, transporting the regular Borel measure $\nu$ through $R_g$ preserves regularity. The measure is nonzero because $Gg=G$, and hence
\begin{align*}
\nu_g(G)=\nu(Gg)=\nu(G)=1.
\end{align*}
Thus $\nu_g$ is a nonzero left-invariant regular Borel measure on $G$, which is exactly a left Haar measure.
[/guided]
[/step]
[step:Use uniqueness of left Haar measure to compare the two measures]
Both $\nu_g$ and $\nu$ are left Haar measures on the locally compact group $G$. By the uniqueness theorem for left Haar measures up to positive scalar (citing a result not yet in the wiki: Uniqueness of left Haar measure up to scalar), there exists a constant $c(g)>0$ such that
\begin{align*}
\nu_g(A)=c(g)\nu(A)
\end{align*}
for every $A\in\mathcal B(G)$.
[/step]
[step:Evaluate on the whole group to force the scalar to be one]
Taking $A=G$ in the preceding identity gives
\begin{align*}
\nu_g(G)=c(g)\nu(G).
\end{align*}
Since $Gg=G$ and $\nu(G)=1$, the left-hand side satisfies
\begin{align*}
\nu_g(G)=\nu(Gg)=\nu(G)=1.
\end{align*}
Thus
\begin{align*}
1=c(g)\cdot 1,
\end{align*}
so $c(g)=1$.
[/step]
[step:Conclude right invariance]
Since the fixed element $g\in G$ was arbitrary, the previous steps show that for every $g\in G$ and every $A\in\mathcal B(G)$,
\begin{align*}
\nu(Ag)=\nu_g(A)=\nu(A).
\end{align*}
Therefore $\nu$ is invariant under every right translation of $G$, so $\nu$ is right invariant.
[/step]
