Now, we shall generalize the well-known Caratheodory extension theorem in classical measure theory for lattice-valued fuzzy measure.

Proof.
(i) It follows from Proposition 3.7.

(ii) Since m* is a lattice-valued fuzzy outer measure, we have
(3.3)m*(μE)≤m(μE).
For given ε>0, there exists (μEn;n=1,2,…) such that ⋁n=1∞(m(μEn))≤m*(μE)+ε [29]. Since μE=μE∧(⋁n=1∞μEn)=⋁n=1∞(μE∧μEn) and by the monotonicity and σ-additivity of m, we have m(μE)≤⋁n=1∞m(μE∧μEn)≤⋁n=1∞m(μEn)≤m*(μE)+ε. Since ε>0 is arbitrary, we conclude that
(3.4)m(μE)≤m*(μE).

From (3.3) and (3.4), m(μE)=m*(μE) is obtained.

(iii) Let μE∈σ(L). In order to prove μE is lattice fuzzy measurable, it suffices to show that
(3.5)m*(μA)≥m*(μA∧μE)+m*(μA∧μEc), for μA≤μE.

For given ε>0, there exists μAn∈σ(L), 1≤n<∞ such that
(3.6)⋁n=1∞m(μAn)≤m*(μA)+ε, μA≤⋁n=1∞(μAn).
Now,
(3.7)μA∧μE≤⋁n=1∞(μAn∧μE),μA∧μEc≤⋁n=1∞(μAn∧μEc).
Therefore,
(3.8)m*(μA∧μE)≤⋁n=1∞m(μAn∧μE),m*(μA∧μEc)≤⋁n=1∞m(μAn∧μEc).
From inequalities (3.6) and (3.8), the inequality (3.5) follows.

(iv) Let m- be the restriction of m* to the m* lattice-valued measurable sets, when we write m-=m*/σ(L-). Now, we must show that σ(L-) is a lattice fuzzy σ-algebra containing σ(L) and m- is a lattice-valued fuzzy measure on σ(L). We show it step by stepin the following.

Step 1. If μA,μB∈σ(L-), then μA∨μB∈σ(L-). It also implies that
(3.9)m*(μE)=m*(μE∧μB)+m*(μE∧μBc).
If we write μE∧μA instead of μE in (3.9),
(3.10)m*(μE∧μA)=m*(μE∧μA∧μB)+m*(μE∧μA∧μBc)
is obtained. Now, if we write μE∧μAc instead of μE in (3.9),
(3.11)m*(μE∧μAc)=m*(μE∧μAc∧μB)+m*(μE∧μAc∧μBc)
is obtained. If we aggregate with (3.10) and (3.11); we have
(3.12)m*(μE)=m*(μE∧μA∧μB)+m*(μE∧μA∧μBc)+m*(μE∧μAc∧μB) +m*(μE∧μAc∧μBc).
If we write μE∧(μA∨μB) instead of μE in (3.12), then we get
(3.13) m*(μE∧(μA∨μB))=m*(μE∧(μA∨μB)∧μA∧μB)+m*(μE∧(μA∨μB)∧(μAc∧μB)) +m*(μE∧(μA∨μB)∧μA∧μBc)+m*(μE∧(μA∨μB)∧μAc∧μBc)=m*(μE∧μA∧μB)+m*(μE∧μAc∧μB)+m*(μE∧μA∧μBc)+m*(L0)=m*(μE∧μA∧μB)+m*(μE∧μAc∧μB)+m*(μE∧μA∧μBc).
From (3.12) and (3.13), we obtain
(3.14)m*(μE)=m*(μE∧(μA∨μB))+m*(μE∧(μA∨μB)c).

Step 2. If μA∈σ(L-), then μAC∈σ(L-). If we write μAc instead of μA in the equality
(3.15)m*(μE)=m*(μE∧μA)+m*(μE∧μAc),
we have
(3.16)m*(μE)=m*(μE∧μAc)+m*(μE∧(μAc)c);(μAc)c=μA=m*(μE∧μAc)+m*(μE∧μA)=m*(μE).
Therefore, it follows that μAC∈σ(L-). Therefore, we showed that σ(L-) is the algebra of lattice sets.

Step 3. Let μA,μB∈σ(L) and μA∧μB=∅, From (3.13), we have
(3.17)m*(μE∧(μA∨μB))=m*(μE∧μAc∧μB)+m*(μE∧μA∧μBc) =m*(μE∧μB)+m*(μE∧μA).Step 4. σ(L-) is a lattice σ-algebra.

From the previous step, we have for every family of (for each disjoint lattice sets) (μBn),n=1,2,…,
(3.18)m*(μE∧(⋁n=1kμBn))=⋁n=1km*(μE∧μBn).

Let μA=⋁n=1∞μAn and μAn∈σ(L). Then, μA=⋁n=1∞μBn, μBn=(μAn∧(⋁x=1n-1μAx)c), and μBi∧μBj=∅ for i≠j. Therefore, we obtain following inequality:
(3.19)m*(μE)≥m*(μE∧(⋁n=1∞μBn))+m*(μE∧(⋁n=1∞μBn)c).
Hence, m* is a lattice σ-semiadditive.

Since σ(L-) is an algebra, ⋁n=1kμBn∈σ(L-) for all n∈N. The following inequality is satisfied for all n:
(3.20)m*(μE)≥m*(μE∧(⋁n=1kμBn))+m*(μE∧(⋁n=1kμBn)c).
From the inequality μE∧(⋁n=1∞μBn)c≤μE∧(⋁n=1∞μBn)c and monotonicity of lattice-valued fuzzy measure and (3.20), we have
(3.21)m*(μE)≥⋁j=1nm*(μE∧μBj)+m*(μE∧μAc).
Then, taking the limit of both sides, we get
(3.22)m*(μE)≥⋁j=1∞m*(μE∧μBj)+m*(μE∧μAc).
Using the semiadditivity, we have,
(3.23)m*(μE∧μA)=m*(⋁j=1∞(μE∧μBj))=m*(μE∧(⋁j=1∞μBj))≤m*(μE∧μBj).
From (3.22), we have
(3.24)m*(μE)≥m*(μE∧μA)+m*(μE∧μAc).
Hence, μA∈σ(L-). This shows that σ(L-) is a lattice fuzzy σ-algebra.

Step 5. m-=m*/σ(L-) is a lattice fuzzy measure, where we only need to show lattice is σ-additive.

Let μE=⋁j=1∞μAj. From (3.22), we have
(3.25)m*(⋁j=1∞μAj)≥⋁j=1∞m*(μAj).

Step 6. We have σ(L-)⊃σ(L).

Let μA∈σ(L) and μE≤μA. Then, we must show the following inequality:
(3.26)m*(μE)≥m*(μE∧μA)+m*(μE∧μAc).

If μE∈σ(L), then μE∧μA and μE∧μAc are different and both of them belong to σ(L), (3.26) is obvious and since m*=m, hence additive.

With μE≤μX and given ε>0, σ(L), there is μEj which contains σ(L) such that we have
(3.27)m*(μE)+ε>⋁j=1∞m(μEj).

Now, from the equality
(3.28)μEj=(μEj∧μA)∨(μEj∧μAc)
and from the Definition 2.1 and Theorem 2.3, we have the following equality:
(3.29)m(μEj)=m(μEj∧μA)+m(μEj∧μAc).

Therefore, we obtain the following:
(3.30)μE∧μA≤⋁j=1∞(μEj∧μA),μE∧μAc≤⋁j=1∞(μEj∧μAc).
Using the monotonicity and semiadditivity, we obtain
(3.31)m*(μE∧μA)≤⋁j=1∞m(μEj∧μA),m*(μE∧μAc)≤⋁j=1∞m(μEj∧μAc).
Using the sum of the inequalities (3.31),
(3.32)m*(μE∧μA)+m*(μEj∧μAc)≤⋁j=1∞m*(μEj)<m*(μE)+ε
is obtained. For arbitrary ε>0, (3.26) is proven. Therefore, (iv) it is obtained as required.

(v) Let σ(L-) be the smallest σ-algebra which contain the σ(L) and let m1 be a lattice fuzzy measure on σ(L-). Then, m1(μE)=m(μE) for all μE∈σ(L). We must show that
(3.33)m1(μA)=m¯(μA).
Since m is a finite σ-lattice fuzzy measure, we can write
(3.34)X=⋁n=1∞μEn, μEn∈σ(L), n≠k, μEn∧μEk=∅, m(μEn)<∞; 1≤n<∞.
If μA∈σ(L-), then we have
(3.35)m¯(μB)=⋁n=1∞m¯(μA∧μEn), m1(μA)=⋁n=1∞m1(μA∧μEn).
To prove the inequality (3.33), it suffices to show that
(3.36)m1(μA)=m¯(μA), μA∈σ(L-), m¯(μA)<∞.

Let μA∈σ(L¯), m¯(μA)<∞, and ε>0 arbitrary. Then, we have
(3.37)μA≤⋁n=1∞μEn, for μEn∈σ(L), 1≤n<∞,(3.38)m¯(⋁n=1∞μEn)≤⋁n=1∞m(μEn)<m¯(μA)+ε.
Since m1(μA)≤m1(⋁n=1∞μEn)≤⋁n=1∞m1(μEn)=⋁n=1∞m(μEn) and from (3.38), we get
(3.39)m1(μA)≤m¯(μA).
Also, from (3.38), we can write μF=∨n=1∞μEn∈σ(L¯) for the sets μEn. Therefore, μF is m* lattice fuzzy measurable. From the inequality μA≤μF and (3.38),
(3.40)m¯(μF)=m¯(μA)+m¯(μF-μA),m¯(μF-μA)=m¯(μF)-m¯(μA)<ε
are obtained.

From the equalities m(μE)=m¯(μE) and m1(μF)=m¯(μF) for all μA∈σ(L), we can write
(3.41)m(μA)≤m(μF)=m1(μF)=m1(μA)+m1(μF-μA)≤m1(μA)+m¯(μF-μA).
Therefore, from (3.41),
(3.42)m¯(μA)≤m1(μA)
is obtained.

Finally from the inequalities (3.41) and (3.39), hence the proof is completed.