Haar measure on an abelian group is invariant under inversion

(Fubinito for continuous compact kernels.) Let $\mathrm{\Lambda },M$ be positive real-linear functionals on $C_c(G;\mathbb{R})$ and let $F\in C_c(G\times G;\mathbb{R})$. Then ${\mathrm{\Lambda }}_x{M}_{y}F=M_y{\mathrm{\Lambda }}_{x}F$. Indeed, let $K_X,K_Y$ be the compact projections of $\mathrm{supp}F$; if $F=0$ both sides vanish, so assume otherwise and choose by [F3] cutoffs $c_X,c_Y\in C_c$ with $0\le c_i\le 1$, $c_X=1$ on $K_X$, $c_Y=1$ on $K_Y$; put $L_Y=\mathrm{supp}{c}_Y$ and $L_X=\mathrm{supp}{c}_X$. For $\epsilon >0$, consider all pairs $(y,V)$ with $y\in L_Y$, $V$ open containing $y$, and $\mathrm{\mid }F(x,y^{\mathrm{\prime }})-F(x,y)\mathrm{\mid }\le \epsilon$ for every $x\in G$, $y^{\mathrm{\prime }}\in V$. Joint continuity and compactness of $K_X$ give such a $V$ at every $y$: a finite cover in the $x$ coordinate gives uniform control on $K_X$, and both sections vanish off $K_X$. The full family of these $V$ covers $L_Y$, so compactness gives finitely many pairs $(y_j,V_j)$ covering $L_Y$; by [F3] choose nonnegative ${\psi }_j\in C_c$ supported in $V_j$ with ${\sum }_j{\psi }_j=1$ on $L_Y$, and put $H(x,y):=c_Y(y){\sum }_j{\psi }_j(y)F(x,y_j)={\sum }_j{c}_Y(y){\psi }_j(y)F(x,y_j)$. Each $F(\cdot ,y_j)$ lies in $C_c(G)$ and each $c_Y{\psi }_j$ lies in $C_c(G)$, so $H$ is a finite sum of products ${\xi }_j(x){\eta }_j(y)$; for such sums the two iterated integrals are equal by linearity and factorization. Moreover $\mathrm{\mid }F-H\mathrm{\mid }\le \epsilon c_X{c}_Y$: on $K_X\times L_Y$ the pointwise convex-combination bound holds, while off $L_Y$ both terms vanish and off $K_X$ both vanish. Applying [F4] twice gives $\mathrm{\mid }{\mathrm{\Lambda }}_x{M}_y(F-H)\mathrm{\mid }\le {\mathrm{\Lambda }}_x{M}_y\mathrm{\mid }F-H\mathrm{\mid }\le \epsilon \mathrm{\Lambda }(c_X)M(c_Y)$ and the same with the order interchanged, so $\mathrm{\mid }{\mathrm{\Lambda }}_x{M}_{y}F-M_y{\mathrm{\Lambda }}_{x}F\mathrm{\mid }\le 2\epsilon \mathrm{\Lambda }(c_X)M(c_Y)$; as $\epsilon >0$ is arbitrary and the cutoff integrals are finite, the two iterated integrals agree.

(Comparison identity.) Let $f\in C_c(G;\mathbb{R})$ and let $u\in C_c(G;\mathbb{R})$ satisfy $u(-x)=u(x)$. Then $I(f)J(u)=J_y{I}_z(f(y+z)u(z))$. To see this, start from the trivial factorization $I(f)J(u)=J_y(I_x(f(x)u(y)))$, replace $u(y)$ by $u(y-x)$ inside the $y$-integral using the translation invariance of $J$ ([F2]), move $I_x$ through $J_y$ by step 1.1 applied to the kernel $(x,y)\mapsto f(x)u(y-x)$, and then substitute $x=y+z$ in the inner $I$-integral using the translation invariance of $I$ ([F1]) to obtain $I_x(f(x)u(y-x))=I_z(f(y+z)u(-z))=I_z(f(y+z)u(z))$; the last equality is the symmetry of $u$. The kernel $f(x)u(y-x)$ is continuous and supported in $\mathrm{supp}f\times (\mathrm{supp}f+\mathrm{supp}u)$, which is compact by [F5].

(The approximating net and its ratios.) Let $D$ be the set of pairs $(E,u)$ where $E$ is a symmetric open neighbourhood of $0$ and $u\in C_c(G;\mathbb{R})$ satisfies $u\ge 0$, $u(-x)=u(x)$, $u(0)=1$, $\mathrm{supp}u\subseteq E$; order $D$ by $(E,u)⪯(E^{\mathrm{\prime }},u^{\mathrm{\prime }})$ when $E^{\mathrm{\prime }}\subseteq E$. This is a directed set: given $(E_1,u_1)$ and $(E_2,u_2)$, [F3] applied to $\{0\}\subseteq E_1\cap E_2$ gives $v\in C_c$ with $v(0)=1$ and $\mathrm{supp}v\subseteq E_1\cap E_2$, and $u_3(x):=v(x)v(-x)$ is symmetric, nonnegative, equals $1$ at $0$ and is supported in $E_1\cap E_2$, so $(E_3,u_3)⪰(E_1,u_1),(E_2,u_2)$. For $(E,u)\in D$ both $I(u)$ and $J(u)$ are positive by [F1], [F2] and the positivity on nonempty open sets, so ${\gamma }_{(E,u)}:=I(u)\mathrm{/}J(u)>0$ is well defined. Fix $f\in C_c(G;\mathbb{R})$ with $f\ge 0$, $f\mathrm{\ne }0$, and put ${\eta }_E:={\sup }_{z\in E}{J}_y(\mathrm{\mid }f(y+z)-f(y)\mathrm{\mid })$ and ${\theta }_E:={\eta }_E\mathrm{/}J(f)$. By step 2.1, $I(f)J(u)=J_y{I}_z(f(y+z)u(z))$, while the factorization $I(u)J(f)=J_y{I}_z(f(y)u(z))$ holds by linearity; subtracting and dividing by $J(u)>0$ gives $J(u)(I(f)-{\gamma }_{(E,u)}J(f))=J_y{I}_z((f(y+z)-f(y))u(z))$. Bounding the right side with [F4], swapping the two integrals by step 1.1 applied to the nonnegative continuous compactly supported kernel $\mathrm{\mid }f(y+z)-f(y)\mathrm{\mid }u(z)$, and using $\mathrm{supp}u\subseteq E$ yields $J(u)\mid I(f)-{\gamma }_{(E,u)}J(f)\mid \le {\eta }_{E}I(u)={\eta }_E{\gamma }_{(E,u)}J(u)$, that is $\mid c-{\gamma }_{(E,u)}\mid \le {\theta }_E{\gamma }_{(E,u)},c:=I(f)\mathrm{/}J(f)>0.$

(Uniform continuity and the ratio limit.) Fix a compact symmetric identity neighbourhood $W$ and put $\tilde{K}=\mathrm{supp}f+W$, compact by [F5]. For $\epsilon >0$ consider all triples $(a,V,W^{\mathrm{\prime }})$ with $a\in \tilde{K}$, $V,W^{\mathrm{\prime }}$ open identity neighbourhoods, $W^{\mathrm{\prime }}$ symmetric and contained in $W$, $W^{\mathrm{\prime }}+W^{\mathrm{\prime }}\subseteq V$, and $\mathrm{\mid }f(a+v)-f(a)\mathrm{\mid }<\epsilon \mathrm{/}3$ for all $v\in V$. Continuity gives such triples at each $a$, so their open sets $a+W^{\mathrm{\prime }}$ cover $\tilde{K}$. Take a finite subcover and put $E={\bigcap }_j{W}_j^{\mathrm{\prime }}$. If $y\in \tilde{K}$, choose $j$ with $y\in a_j+W_j^{\mathrm{\prime }}$; for $z\in E$, both $y$ and $y+z$ lie in $a_j+V_j$, so $\mathrm{\mid }f(y+z)-f(y)\mathrm{\mid }<2\epsilon \mathrm{/}3$. If $y\mathrm{\notin }\tilde{K}$, both values vanish, because $z\in E\subseteq W$ and $W=-W$. Thus the difference is supported in $\tilde{K}$ and bounded by $\epsilon$, giving ${\eta }_E\le \epsilon \nu (\tilde{K})$. Hence ${\eta }_E\to 0$ as $E$ shrinks. For the fixed nonzero $f$ of step 3.1, choose $E_0$ with ${\theta }_{E_0}<1\mathrm{/}2$. For every later pair $(E,u)$, $E\subseteq E_0$ implies ${\theta }_E\le {\theta }_{E_0}$, and the inequality of step 3.1 gives $c\mathrm{/}(1+{\theta }_{E_0})\le {\gamma }_{(E,u)}\le c\mathrm{/}(1-{\theta }_{E_0})$. The length of this interval tends to zero as $E_0$ shrinks, so the ratio net is Cauchy and converges by [F6]. It is eventually bounded below by $c\mathrm{/}2>0$, so its limit $\gamma$ is positive. All covers used the complete families of admissible neighborhoods and only finite subfamilies, without uncountable selections.

(Passing to the limit.) Fix $g\in C_c(G;\mathbb{R})$ with $g\ge 0$ and let $\delta >0$. By the uniform continuity argument of step 4.1 applied to $g$ there is a symmetric open $E_0$ with ${\eta }_{E_0}(g):={\sup }_{z\in E_0}{J}_y\mathrm{\mid }g(y+z)-g(y)\mathrm{\mid }\le \delta$. For every $\lambda =(E,u)⪰(E_0,\cdot )$ one has $\mathrm{supp}u\subseteq E\subseteq E_0$, so step 3.1 with the pair $(g,u)$ gives $\mathrm{\mid }I(g)-{\gamma }_{\lambda }J(g)\mathrm{\mid }\le {\eta }_{E_0}(g){\gamma }_{\lambda }\le \delta {\gamma }_{\lambda }$. Letting $\lambda$ run through $D$ and using ${\gamma }_{\lambda }\to \gamma$ gives $\mathrm{\mid }I(g)-\gamma J(g)\mathrm{\mid }\le \delta \gamma$ for every $\delta >0$, so $I(g)=\gamma J(g)$; by linearity of both functionals the identity $I(h)=\gamma J(h)$ holds for every $h\in C_c(G;\mathbb{R})$.

(The scale is one.) By step 5.1 the two Radon measures $m$ and $\gamma \nu$ have equal integrals of every real $C_c$ function, so [F6] gives $m(E)=\gamma \nu (E)$ for every Borel $E$, that is $m(E)=\gamma m(-E)$. Replacing $E$ by $-E$ gives $m(-E)=\gamma m(E)$, hence $m(E)={\gamma }^{2}m(E)$ for every Borel $E$. Choosing a compact neighbourhood $E$ of $0$, [F1] gives $0<m(E)<+\mathrm{\infty }$, so ${\gamma }^2=1$ and, since $\gamma >0$, $\gamma =1$. Therefore $m(-E)=m(E)$ for every Borel set $E$.