In this section, we study coupled random coincidence and coupled random fixed point theorems for a pair of random mappings F:Ω×(X×X)→X and g:Ω×X→X. Then we will prove some results for random mixed monotone mappings, which are the extensions of corresponding results for deterministic mixed monotone mappings of Karapınar et al. [17].

Proof.
Let Θ={η:Ω→X} be a family of measurable mappings. Define a function h:Ω×X→ℝ+ as h(ω,x)=d(x,g(ω,x)). Since x→g(ω,x) is continuous, for all ω∈Ω, we conclude that h(ω,·) is continuous, for all ω∈Ω. Also, since ω→g(ω,x) is measurable, for all x∈X, we conclude that h(·,x) is measurable, for all x∈X (see [23, page 868]). Thus, h(ω,x) is the Caratheodory function. Thus, if η:Ω→X is measurable mapping, then ω→h(ω,η(ω)) is also measurable (see [24]). Also, for each θ∈Θ, the function η:Ω→X defined by η(ω)=g(ω,θ(ω)) is measurable; that is, η∈Θ.

Now we are going to construct two sequences of measurable mappings {ξn} and {ηn} in Θ and two sequences {g(ω,ξn(ω))} and {g(ω,ηn(ω))} in X as follows. Let ξ0,η0∈Θ be such that g(ω,ξ0(ω))≤F(ω,(ξ0(ω),η0(ω))) and g(ω,η0(ω))≥F(ω,(η0(ω),ξ0(ω))), for all ω∈Ω. Since F(ω,(ξ0(ω),η0(ω)))∈F(ω×(X×X))⊆g(ω×X), by a sort of Filippov measurable implicit function theorem (see [25, 26]), there is ξ1∈Θ such that g(ω,ξ1(ω))=F(ω,(ξ0(ω),η0(ω))). Similarly, as F(ω,(η0(ω),ξ0(ω)))∈g(ω×X), there is η1∈Θ such that g(ω,η1(ω))=F(ω,(η0(ω),ξ0(ω))). Thus F(ω,(ξ0(ω),η0(ω))) and F(ω,(η0(ω),ξ0(ω))) are well defined now. Again, since
(7)F(ω,(ξ1(ω),η1(ω))),F(ω,(η1(ω),ξ1(ω)))∈g(ω×X),
there are ξ2,η2∈Θ such that
(8)g(ω,ξ2(ω))=F(ω,(ξ1(ω),η1(ω))),g(ω,η2(ω))=F(ω,(η1(ω),ξ1(ω))).
Continuing this process we can construct sequences {ξn(ω)} and {ηn(ω)} in X such that
(9)g(ω,ξn+1(ω))=F(ω,(ξn(ω),ηn(ω))),g(ω,ηn+1(ω))=F(ω,(ηn(ω),ξn(ω))),
for all n∈ℕ. Now, we use mathematical induction to prove that
(10)g(ω,ξn(ω))≤g(ω,ξn+1(ω)),g(ω,ηn(ω))≥g(ω,ηn+1(ω)),
for all n∈ℕ. Let n=0, and by assumption we have
(11)g(ω,ξ0(ω))≤F(ω,(ξ0(ω),η0(ω))),g(ω,η0(ω))≥F(ω,(η0(ω),ξ0(ω))).
Since
(12)g(ω,ξ1(ω))=F(ω,(ξ0(ω),η0(ω))),g(ω,η1(ω))=F(ω,(η0(ω),ξ0(ω))),
we have
(13)g(ω,ξ0(ω))≤g(ω,ξ1(ω)),g(ω,η0(ω))≥g(ω,η1(ω)).
Therefore, (10) holds for n=0. Suppose (10) holds for some fixed number n≥0. Then, since
(14)g(ω,ξn(ω))≤g(ω,ξn+1(ω)),g(ω,ηn(ω))≥g(ω,ηn+1(ω))
and F is monotone g-nondecreasing in its first argument, we have
(15)F(ω,(ξn(ω),ηn(ω)))≤F(ω,(ξn+1(ω),ηn(ω))),F(ω,(ηn(ω),ξn(ω)))≥F(ω,(ηn+1(ω),ξn(ω))).
Also, since g(ω,ξn(ω))≤g(ω,ξn+1(ω)) and g(ω,ηn(ω))≥g(ω,ηn+1(ω)). and F is monotone g-nonincreasing in its second argument, we have
(16)F(ω,(ξn+1(ω),ηn+1(ω)))≥F(ω,(ξn+1(ω),ηn(ω)))F(ω,(ηn+1(ω),ξn(ω)))≥F(ω,(ηn+1(ω),ξn+1(ω))).
Thus, from (9), we get
(17)g(ω,ξn+1(ω))≤g(ω,ξn+2(ω)),g(ω,ηn+1(ω))≥g(ω,ηn+2(ω)).
Thus, by mathematical induction, we conclude that (10) holds for all n∈ℕ. Now, we prove that {g(ω,ξn(ω))} and {g(ω,ηn(ω))} are Cauchy sequences. Let n∈ℕ, and, by (6)–(10), we have
(18)d(F(ω,(ξn(ω),ηn(ω))),F(ω,(ξn-1(ω),ηn-1(ω)))) ≤φ(max{d(g(ω,ξn(ω)),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiid(g(ω,ηn(ω)),g(ω,ηn-1(ω)))}) +Lmin{d(F(ω,(ξn(ω),ηn(ω))),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiiid(F(ω,(ξn-1(ω),ηn-1(ω))),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiid(F(ω,(ξn(ω),ηn(ω))),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiid(f(ω,(ξn(ω),ηn(ω))),g(ω,ξn-1(ω)))},
which implies that
(19)d(g(ω,ξn+1(ω)),g(ω,ξn(ω)))iiii≤φ(max{d(g(ω,ξn(ω)),g(ω,ξn-1(ω))),iiiiiiiiiiiiiiiiiiiid(g(ω,ηn(ω)),g(ω,ηn-1(ω)))}).
Similarly, we have
(20)d(F(ω,(ηn(ω),ξn(ω))),F(ω,(ηn-1(ω),ξn-1(ω)))) ≤φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}) +Lmin{d(F(ω,(ηn(ω),ξn(ω))),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ηn-1(ω),ξn-1(ω))),g(ω,ηn(ω))),iiiiiiiiiiiiiiiiid(F(ω,(ηn(ω),ξn(ω))),g(ω,ηn(ω))),iiiiiiiiiiiiiiiid(F(ω,(ηn-1(ω),ξn-1(ω))),g(ω,ηn-1(ω)))},
which implies that
(21)d(g(ω,ηn+1(ω)),g(ω,ηn(ω))) ≤φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}).
From (19) and (21), we get that
(22)max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))} ≤φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}).
Since φ(t)<t, for all t>0, by (22), we have
(23)max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))} <max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))}.
Set dn=max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),d(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}, then {dn} is a nonincreasing sequence of positive real numbers. Thus, there is d≥0 such that
(24)limn→∞dn=d.

Suppose that d>0; letting n→∞ in two sides of (22) and using the properties of φ, we have
(25)d=limn→∞dnii≤limn→∞φ(max{d(g(ω,ηn(ω)),g(ω,ηn-1(ω))),iiiiiiiiiiiiiiiiiiiiiiiiiid(g(ω,ξn(ω)),g(ω,ξn-1(ω)))})ii≤φ(d)<d,
which is a contradiction. Hence d=0; that is,
(26)d=limn→∞max{d(g(ω,ξn+1(ω)),g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiiiiid(g(ω,ηn+1(ω)),g(ω,ηn(ω)))}=0.
We will show that {g(ω,ηn(ω))} and {g(ω,ξn(ω))} are Cauchy sequences. Suppose, to the contrary, that at least one of {g(ω,ηn(ω))} or {g(ω,ξn(ω))} is not a Cauchy sequence. This means that there exists an ɛ>0 for which we can find subsequences {g(ω,ηn(k)(ω))} of {g(ω,ηn(ω))} and {g(ω,ξn(k)(ω))} of {g(ω,ξn(ω))} with n(k)>m(k)≥k (k=1,2,…) such that
(27)max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}≥ɛ.
Further, corresponding to m(k), we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k)≥k and satisfies (27). Then,
(28)max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω))),iiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω)))}<ɛ.

Using the triangle inequality and (28), we have
(29)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))) ≤d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))) +d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω))) ≤d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))+ɛ,d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω))) ≤d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)) +d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω))) ≤d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))+ɛ.
By (27) and (29), we obtain
(30)ɛ≤max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}i≤max{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}+ɛ.
Letting k→∞, in the inequalities above, we get
(31)limk→∞max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}=ɛ.
By the triangle inequalities, we have
(32)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))) ≤d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))) +d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))) +d(g(ω,ξm(k)-1(ω)),g(ω,ξm(k)(ω))),d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω))) ≤d(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω))) +d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω))) +d(g(ω,ηm(k)-1(ω)),g(ω,ηm(k)(ω))).

By the above inequalities and (27), we have
(33)ɛ≤max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))}i≤max{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))} +max{d(g(ω,ξn(k)-1(ω)),iiiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω)) d(g(ω,ηn(k)-1(ω)),iiiiiiiiiiiiiiiiiiig(ω,ηm(k)-1(ω)))} +max{d(g(ω,ξm(k)-1(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηm(k)-1(ω)),g(ω,ηm(k)(ω)))}.
Again, by the triangle inequality, we obtain
(34)d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))) ≤d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)(ω))) +d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω))) ≤d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω)))+ɛ,d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω))) ≤d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)(ω))) +d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω))) ≤d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω)))+ɛ.
Therefore,
(35)max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}iiiiiiii≤max{d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiiiiid(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω)))}+ɛ.
Taking k→∞ in (33) and (35), we have
(36)limn→∞max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}=ɛ.
Since n(k)>m(k), g(ω,ξn(k)-1(ω))≥g(ω,ξm(k)-1(ω)) and g(ω,ηn(k)-1(ω))≤g(ω,ηm(k)-1(ω)). Then, from (6)–(10), we get
(37)d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))) =d(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiF(ω,(ξm(k)-1(ω),ηm(k)-1(ω)))) ≤φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}) +Lmin{d(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξm(k)-1(ω),ηm(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξn(k)-1(ω),ηn(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiiid(F(ω,(ξm(k)-1(ω),ηm(k)-1(ω))),iiiiiiiiiiiiiiiiiig(ω,ξm(k)-1(ω)))} ≤φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}) +Lmin{d(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω)))}.
Similarly,
(38)d(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω))) ≤φ(max{d(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}) +Lmin{d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}.
From (37) and (38), we arrive at
(39)max{d(g(ω,ξn(k)(ω)),g(ω,ξm(k)(ω))),iiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηm(k)(ω)))} ≤φ(max{d(g(ω,ξn(k)-1(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiid(g(ω,ηn(k)-1(ω)),g(ω,ηm(k)-1(ω)))}) +Lmin{d(g(ω,ξm(k)(ω)),g(ω,ξm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ξn(k)(ω)),g(ω,ξn(k)-1(ω)))} +Lmin{d(g(ω,ηm(k)(ω)),g(ω,ηm(k)-1(ω))),iiiiiiiiiiiiiiiiid(g(ω,ηn(k)(ω)),g(ω,ηn(k)-1(ω)))}.
Letting n→∞ in the above inequality and using (26), (27), and the properties of φ, we have
(40)ɛ≤φ(ɛ)+2Lmin{0,0}<ɛ,
which is a contradiction. This means that {g(ω,ξn(ω))} and {g(ω,ηn(ω))} are Cauchy sequences.

Since X is complete, for all ω∈Ω, there exist the functions ζ(ω) and θ(ω) such that
(41)limn→∞g(ω,ξn(ω))=ζ(ω), limn→∞g(ω,ηn(ω))=θ(ω).
Thus,
(42)limn→∞F(ω,(ξn(ω),ηn(ω)))=limn→∞g(ω,ξn(ω))=ζ(ω),limn→∞F(ω,(ηn(ω),ξn(ω)))=limn→∞g(ω,ηn(ω))=θ(ω).

Since F and g are compatible mappings, we have
(43)limn→∞d(g(ω,F(ω,(ξn(ω),ηn(ω)))),iiiiiiiiiiiiF(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))=0,limn→∞d(g(ω,F(ω,(ηn(ω),ξn(ω)))),iiiiiiiiiiiiF(ω,(g(ω,ηn(ω)),g(ω,ξn(ω)))))=0.

Suppose at first that assumption (a) holds. Taking the limit as n→∞ in the following inequalities
(44)d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω))))) ≤d(g(ω,ζ(ω)),g(ω,F(ω,(ξn(ω),ηn(ω))))) +d(g(ω,F(ω,(ξn(ω),ηn(ω)))),iiiiiiiiiiiiiF(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))))
and using (9) and the continuity of F, g, we get
(45)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω))))=θ.
This implies g(ω,ζ(ω))=F(ω,(ζ(ω),θ(ω))). Similarly, we can show that g(ω,θ(ω))=F(ω,(θ(ω),ζ(ω))) for each ω∈Ω. The proof is complete.

Suppose now that (b) holds. From (9), we have
(46)limn→∞F(ω,g(ω,ξn(ω)),g(ω,ηn(ω))) =limn→∞g(ω,F(ω,(ξn(ω),ηn(ω)))) =limn→∞g(ω,g(ω,ξn+1(ω))) =g(ω,ζ(ω)),(47)limn→∞F(ω,g(ω,ηn(ω)),g(ω,ξn(ω))) =limn→∞g(ω,F(ω,(ηn(ω),ξn(ω)))) =limn→∞g(ω,g(ω,ηn+1(ω))) =g(ω,θ(ω)).
Since g(ω,g(ω,ξn(ω)))≤g(ω,ζ(ω)) and g(ω,g(ω,ηn(ω)))≥g(ω,θ(ω)), we have
(48)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))) ≤d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω))))) +d(F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω)))),iiiiiiiiiiiiF(ω,(ζ(ω),θ(ω)))((ξn(ω)))) ≤d(g(ω,ζ(ω)),F(ω,(g(ω,ξn(ω)),g(ω,ηn(ω))))) +φ(max{d(g(ω,g(ω,ξn(ω))),g(ω,ζ(ω))),iiiiiiiiiiiiiiiiiid(g(ω,g(ω,ηn(ω))),g(ω,θ(ω)))}) +Lmin{d(g(ω,g(ω,ξn(ω))),F(ω,(ζ(ω),θ(ω)))),iiiiiiiiiiiiiiiiiid(g(ω,ζ(ω)),iiiiiiiiiiiiiiiiiF(ω,(g(ω,ξn(ω),g(ω,ηn(ω))))),iiiiiiiiiiiiiiiiid(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))),iiiiiiiiiiiiiiiiid(g(ω,g(ω,ξn(ω))),iiiiiiiiiiiiiiiiiF(ω,(g(ω,ξn(ω),g(ω,ηn(ω)))))}.
Taking n→∞ in the above inequality and using (46) and the properties of φ, we have
(49)d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))) ≤φ(max{0,0}) +Lmin{d(g(ω,ζ(ω)),F(ω,(ζ(ω),θ(ω)))),0}=0.
Hence g(ω,ζ(ω))=F(ω,(ζ(ω),θ(ω))).

Similarly, one can show that g(ω,θ(ω))=F(ω,(θ(ω),ζ(ω))).

The proof is complete.