Fixed point theorems and their applications in Measure Theory

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We know that there are many versions for the fixed point theorem and they have many applications.

I would like to know whether there is one has an application in Measure Theory.

And I would be grateful if I was informed good references about it.

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From the Kakutani's fixed point theorem:

Theorem $1$. Let $E$ be a real vector space endowed with a dot product, let $K$ be a nonempty compact convex set of $E$ and let $G$ be a compact subgroup of $\textrm{GL}(E)$. Assume that $G$ stabilises $K$, then there exists $x\in K$ such that for all $g\in G$, $g(x)=x$.

One can deduce the existence of Haar's measures on compact topological groups:

Theorem $2$. Let $G$ be a compact topological group, then there exists a unique Borel probability measure $\mu$ on $G$ such that for all $g\in G$ and all Borel measurable subset $A$ of $G$, one has: $$\mu(gA)=\mu(A).$$ In particular, for all measurable map $f\colon G\rightarrow\mathbb{R}$ and all $g\in G$, one has: $$\int_Gf(gx)\,\mathrm{d}\mu(x)=\int_Gf(x)\,\mathrm{d}\mu(x).$$

Reference. A. Weil. L’intégration dans les groupes topologiques et ses applications. Hermann, 1965.

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The existence of an invariant probability distribution for a Markov process follows naturally from a fixed point argument, indeed an invariant distribution is essentially defined to be a fixed point.

Let $X$ be a compact metrizable space, $\Delta(X)$ be the space of Borel probability measures on $X$ endowed with the weak*-topology induced by identifying it with a subset of $C(X)^*$. Let $k:X\to\Delta(A)$ be continuous. The function $\phi:\Delta(X)\to\Delta(X)$ given by $$\phi(\mu)(A)=\int k(x)(A)~\mathrm d\mu(x)$$ for $A$ each Borel subset of $X$ is a continuous function from a compact convex subset of locally convex Hausdorff topological vector space to itself and has a fixed point by the Schauder–Tychonoff fixed point theorem. Such a fixpoint is exactly an invariant distribution. Any book on random dynamical systems will contain this result, proven in essentially this way.