The Banach fixed point theorem has the following statement
THEOREM ( Banach contraction principle). Let $(Y,d)$ be a complete metric space and $F:Y\to Y$ be contractive . Then $F$ has a uniqe fixed point $u$, and $F^n(y)\to u$ for each $y\in Y$.
This statement is from the excellent book Fixed Point Theory written by Andrzej Granas and James Dugundji. In this book several generalizations of this theorem are given. Several of them are to relax the hypothesis of classic contraction to another contraction hypothesis that is weaker. For example, $d(F (x) ,F (y)) \le \phi(d (x, y))$ for all $x,y\in Y$ with $ \lim_{n \to \infty}\phi^n (d (x, y )) = 0$ ( see p. 15).
QUESTION Let $ (M, d) $ be a complete metric space and $ (M\times M, \mathcal{M}\otimes\mathcal{M}, \mu) $ be a probability space. Suppose that $F:M\to M$ is continuous and a.s. contractive i.e. there is a $0<\lambda<1$ such that $$ \mu\left\{ (x,y) :\quad d(F (x) ,F (y)) \le \lambda \cdot d(x, y)\quad \right\}=1. $$ So is there a counterexample to the claim that $ F $ has a unique fixed point? If not, a reference in the form of a book that treats the subject would be well accepted answer.
My attempt. Let $\delta_{(x, y)} $ the measure of Dirac concentrated in $ (x, y) $. The above question has a positive answer to the dirac measure (equivalent to the deterministic case). I have tried to prove my assertion probability measures of the form $$ \mu=\dfrac{1}{1+\epsilon}\left(\delta_{(x,y)}+\epsilon\cdot \nu \right) $$ for a $ \epsilon>0 $ small enough and any probability measure $\nu$ on $(M\times M, \mathcal{M}\otimes\mathcal{M})$.