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Give $C(G)$, the space of continuous functions on a compact space, its uniform norm. The left regular action is $l_{g}f(x)=f(g^{-1}x)$. The group law gives $l_g{l}_h=l_{gh}$ and $l_e=I$, and each translation is a linear isometry because $x\mapsto g^{-1}x$ is a bijection of $G$.

For fixed $f\in C(G)$, the map $(g,x)\mapsto f(g^{-1}x)$ is continuous. Compactness of the $x$-space makes this dependence uniform: for a given $g_0$ and $\epsilon >0$, continuity gives a neighbourhood of $g_0$ and a neighbourhood of each $x$ on which the difference from $f(g_0^{-1}x)$ is less than $\epsilon$; a finite cover of $G$ permits one common neighbourhood of $g_0$. Hence $\Vert l_{g}f-l_{g_0}f{\Vert }_{\infty }\to 0$. The estimatethen proves the joint continuity of the action on $G\times C(G)$.

For a left action on the continuous dual space, use the contragredient conventionIt satisfies $T_g{T}_h=T_{gh}$ and preserves the dual unit ball $B^{\prime }$. If the printed prime denotes the literal transpose $l_g^{\ast }\varphi =\varphi \circ l_g$, these maps instead satisfy $l_g^{\ast }{l}_h^{\ast }=l_{hg}^{\ast }$ and form a right action; replacing $g$ by $g^{-1}$ gives the left action above. Their invariant functionals are the same.

For $g_i\to g$ and ${\varphi }_i\to \varphi$ in the weak-star topology on $B^{\prime }$, every $f\in C(G)$ satisfiesThe first term uses strong continuity of translation and $\Vert {\varphi }_i\Vert \le 1$; the second uses weak-star convergence at one fixed function. This proves the requested joint weak-star topology continuity.

To obtain Haar measure, consider the set $K$ of positive linear functionals $\varphi$ on $C(G)$ with $\varphi (1)=1$. Such functionals have norm one: for complex $f$, choose a phase making $\varphi (f)$ real and nonnegative and use $\mathrm{Re}(e^{it}f)\le \Vert f{\Vert }_{\infty }$; this gives $|\varphi (f)|\le \Vert f{\Vert }_{\infty }$, with equality on the constant-one function. Thus $K\subseteq B^{\prime }$. Evaluation at the identity belongs to $K$, and positivity and normalization are weak-star closed conditions. The Banach-Alaoglu theorem makes $K$ a nonempty compact convex set. Each $T_g$ is a continuous affine self-map of it.

Here is the fixed-point step, including why no commutativity assumption on $G$ is needed. For any finite list $g_1,\dots ,g_n$, the Schauder-Tychonoff fixed-point theorem gives a fixed point $\varphi \in K$ of $n^{-1}{\sum }_j{T}_{g_j}$. Let $H$ be the closed subgroup they generate and, for real $f\in C(G)$, set $h(x)=\varphi (l_{x}f)$ on $H$. It is continuous and satisfiesAt a maximum point $x_0$, every summand must have the same maximum value. Iteration gives that value at all inverse-generator words times $x_0$. The closure of their positive-word monoid is a closed subsemigroup of a compact group, hence is the entire generated subgroup $H$. Density and continuity make $h$ constant on $H$. Therefore $\varphi$ is fixed by each of $T_{g_1},\dots ,T_{g_n}$, first on real functions and then on complex functions by linearity. The closed fixed-point sets consequently have the finite intersection property in $K$, so compactness gives a common fixed functional for every $g\in G$.

The Riesz-Markov-Kakutani representation theorem converts this positive functional into a regular Borel probability measure ${\mu }_l$ with $\varphi (f)=\int fd{\mu }_l$. Its invariance gives left translation invariance of the measure. It has full support: if a nonempty open set had measure zero, a finite collection of its left translates would cover the compact group and force ${\mu }_l(G)=0$, contradicting normalization. Thus it is a normalized left Haar measure. The right translation action gives a normalized right Haar measure ${\mu }_r$ by the same argument.

For uniqueness and equality, take any $f\in C(G)$ and apply the Fubini theorem toFor fixed $y$, left invariance of ${\mu }_l$ makes the inner integral $\int fd{\mu }_l$. Reversing the order, for fixed $x$, right invariance of ${\mu }_r$ makes the inner integral $\int fd{\mu }_r$. Since both total masses are one,Uniqueness in the Riesz-Markov-Kakutani representation theorem gives $\boxed{{\mu }_l={\mu }_r}$. Comparing any normalized left Haar measure to this fixed right Haar measure proves uniqueness as well. The conclusion is uniqueness of normalized Haar measure. Without normalization, the printed uniqueness claim is false: even on the one-point group, ${\delta }_e$ and $2{\delta }_e$ are distinct Haar measures. In general the result is uniqueness up to a positive scalar, and left and right probability normalizations coincide.