On a locally-compact group $h(x^{-1})$ constant a.e. $\implies h(x)$ constant a.e. (proof of uniqueness of Haar measure)

Question

I'm trying to understand the final step in a proof of uniqueness of left Haar measures on locally-compact groups as presented in Knapp's "Advanced Real Analysis" (pg. 224-225) or this note by Pedersen (pg. 5). Both proofs proceed roughly as follows

Let $\mu,\lambda$ be left Haar measures on the $\sigma$-compact LCG $G$, wlog assume $\mu\leq\lambda$ (otherwise replace $\lambda$ by $\mu+\lambda$). By Radon-Nikodym there exists $h:G\to[0,\infty)$ with $d\mu=hd\lambda$. For $g$ arbitrary fixed and $f\geq0$ arbitrary by translation-invariance (here $L_gf(x)=f(g^{-1}x)$) $$\int f\cdot L_gh\,d\lambda=\int L_{g^{-1}}f\cdot h\,d\lambda=\int L_{g^{-1}}f\,d\mu=\int f\,d\mu=\int f\cdot h\,d\lambda$$ $\implies L_gh\,d\lambda=h\,d\lambda$ as measures $\implies$ for any $g$ holds $L_gh=h$ $\lambda$-a.e.. It follows that the function $g\mapsto\int|L_gh-h|\,d\lambda(x)$ is zero, so by Tonelli $$\iint|L_gh-h|\,d\lambda(g)d\lambda(x)=\iint|L_gh-h|\,d\lambda(x)d\lambda(g)=0,$$ hence for $\lambda$-almost every choice of $x$ holds $$\int|h(g^{-1}x)-h(x)|\,d\lambda(g)=\int|h(g^{-1})-h(x)|\,d\lambda(g)=0.$$ Fix such an $x_0$, then $h(g^{-1})=h(x_0)$ for $\lambda$-almost every $g\implies h(x^{-1})$ is constant $\lambda$-a.e. $\implies h(x)$ is constant $\lambda$-a.e..

I don't understand the highlighted step, how does $h(x^{-1})$ being costant a.e. imply the unmodified $h(x)$ is constant a.e.? Both Knapp and Pedersen gloss over this detail.

Knapp: [...] and $h(g^{-1}x)=h(x)$ for almost every $x\in G$. We can regard $h(g^{-1}x)$ and $h(x)$ as functions of $(g,x)\in G\times G$; for each $g$, they are equal for almost every $x$. By Fubini’s Theorem they are equal for almost every pair $(g, x)$ (with respect to the product measure), and then for almost every $x$ they are equal for almost every $g$. Pick one such $x$, say $x_0$, then it follows that $h(x) = h(x_0)$ for almost every $x$.

Pedersen: Fubini’s theorem, applied to the product integral and the function $(x,z)\mapsto h(z^{-1}x)-h(x)$, now shows that the function $x\mapsto\int|h(z^{-1}x)-h(x)|dz=\int|h(z^{-1})-h(x)|dz$ equals zero almost everywhere. It follows that $h$ is constant almost everywhere, as desired.

Answer 1

The problem here is that you are trying to fix $x_0$ and use $h(x_0)$ as the constant. Instead, argue as follows. Having established that
$$\int|h(g^{-1})-h(x)|\,d\lambda(g)=0\text{,}$$ for a.e. $x$, go back to Tonelli again and conclude $$\int|h(g^{-1})-h(x)|\,d\lambda(x)\,d\lambda(g)=\int|h(g^{-1})-h(x)|\,d\lambda(g)\,d\lambda(x)=0\text{,}$$ so that for almost every every $g$ we have $$\int|h(g^{-1})-h(x)|\,d\lambda(x)=0\text{.}$$ Then fix such a $g=g_0$ and observe that $h(g_0^{-1})=h(x)$ for $\lambda$-a.e. $x$.

Answer 2

Knapp's argument shows that there is a $\lambda$-null set $N\subset G$ such that if $x_0\in N^c=G\backslash N$ and we set $h_{x_0}(g) := |h(g^{-1}x_0)-h(x_0)|$ then $\text{supp}(h_{x_0}) = L_{x_0}$ where $L_{x_0}$ is a $\lambda$-null set. Let $\iota\colon G\to G$ be given by $\iota(g)=g^{-1}$. Then it is easy to check that $h(x)\neq h(x_0)$ if and only if $x \in \iota(M_{x_0})$ where $M_{x_0}:=x_0^{-1}L_{x_0}$, and by left-invariance $M_{x_0}$ is $\lambda$-null. Thus the following claim implies that $h$ is constant a.e. and so completes the proof of uniqueness:

Claim: $\iota(M_{x_0})$ is $\lambda$-null.

Proof: Suppose not, so that $\iota(M_{x_0})$ is not $\lambda$-null. Then as $\lambda(N)=0$ we must have $\iota(M_{x_0})\cap N^c\neq \emptyset$ and so we may pick $x_1 \in\iota(M_{x_0})\cap N^c$. But $x_1 \in N^c$ implies, exactly as for $x_0$, that we must have $h(x)= h(x_1)$ if and only if $x\in \iota(M_{x_1})^c$ where $M_{x_1}$ is $\lambda$-null, and since $x_1 \in \iota(M_{x_0})$, if $x \in \iota(M_{x_1})^c$ then $h(x)=h(x_1)\neq h(x_0)$ and hence $\iota(M_{x_1})^c \subseteq \iota(M_{x_0})$, or equivalently $\iota(M_{x_0})\cup\iota(M_{x_1}) =G$. Applying $\iota$ to this equality $M_{x_0}\cup M_{x_1} = G$ and hence $\lambda(G) =0$, which is a contradiction since $G$ is open. Thus $\iota(M_{x_0})$ must be null and $h$ is constant almost everywhere as required.