Consider now the proof of Theorem 1.4.
Proof.

(a)
If such that , then (1.9) follows directly from (1.1).
Now, fix a point with . To estimate the second term on the right of (1.1), we can apply Jensen's inequality (see [13]), by ():
and therefore
This inequality, together with Definition 1.3(a), implies that the expression
satisfies the inequality
By Lemma 2.3 (b) and Definition 1.3 (b), , and hence (1.9) can be deduced from
The properties defining are trivial if with . So assume such that .
First, we show that the function is almost measurable on . It is an easy consequence of () and Lemma 2.4 that the functions
are almost measurable on . This means that there exists a measurable subset of such that and (3.6), (3.7) are measurable on . Further, since is almost measurable on , it can be supposed that is measurable on . Thus we need to show that the function
is measurable on . To prove this let
The measurability of (3.6) on implies that . Since
it is enough to show that (3.8) is measurable on . It follows from the definitions of and that the function
is measurable. Hence, by Lemma 2.3 (d), (3.8) is measurable on .
It is now clear that is almost measurable on .
Next, we prove that is integrable over .
To prove this, it is enough to show that the function
is bounded on . Since
we need to verify that (3.12) is bounded on . By (), we can find a such that
It therefore follows from
that
for all , and hence
Thus
and therefore another application of (3.14) gives
and from this the claim follows. Consequently, is integrable over , as required.
The result is completely proved.
Now we are in a position to prove Theorem 1.5.
Proof.
We begin with the proof of (c) and (d).

(c)
To prove that we use induction on . Clearly belongs to . Let such that the assertion holds. Then Lemma 2.6 yields that . We show now that the sequence is increasing. By our hypotheses on , it follows that , and we again complete the proof by induction. Suppose such that . Then, by Lemma 2.1 and the induction hypothesis
Since
Theorem 1.4(a) implies that
Because of (3.22) and the fact that is increasing, there exists a function such that converges pointwise on to . Then is a consequence of together with (3.22), Lemma 2.1(a) and, Theorem 1.4(b). If with , then according to the continuity of on , converges to . Let with . As we have seen is increasing, and therefore for all . Consequently
Now (1.11) and the monotone convergence theorem give that is a solution of (1.2).

(d)
We can see exactly as in the proof of (c) that , and the sequence is decreasing. It follows from an easy induction argument that is nonnegative for all . Linking up with the foregoing, there exists a function such that converges pointwise on to . can be shown as in the proof of (c).
The continuity of on implies that converges to for every .
Assume now that . If such that for every large enough , then
If such that for every , then
which leads to . In both cases converges to for every .
According to (1.11) and the monotone convergence theorem is a solution of (1.2).
() For convergence of the successive approximations
to a solution of (1.2) it suffices, in view of (c), to show that is a solution of (1.1), which is evident. It remains to prove that if is a solution of (1.6), then for every . To this end, it is enough to show that
This is true for , since the functions and are nonnegative. Let for which the result holds. Then, because of the nonnegativity of and the fact that is increasing,
and the proof of the induction step is complete.
() Let
Choose . The set of the upper bounds from for the solutions of (1.1) on is denoted by . By Theorem 1.4, is not empty. Let
Then is a minorized subset of (the elements of are nonnegative). Since this space is order complete (see [12]),
exists. Let , and let be a solution of (1.1) such that . , gives that
It follows that
because is increasing. Since the proof of Lemma 2.5 shows that , and hence by (), the function defined by
belongs to , and therefore . It now comes from (3.33) and the definition of that
Since is increasing,
and thus
We can see that
If we set
then (3.38) shows that is a solution of (1.2). (3.33) gives that
for every solutions of (1.1) for which .
Repeat the preceding construction for every . We thus obtain a set of functions , each a solution of (1.2) on its domain. Moreover, if is a solution of (1.1), then for every we have
Introduce the next functions: for every with let the function be defined on by
Obviously, these functions are also solutions of (1.2) on their domains.
Now let , such that and . Using (3.41), it is easy to verify that
and therefore Lemma 2.10 is applicable to the solutions . This gives a unique solution of (1.2) for which
It remains to prove that is maximal. Let be a solution of (1.1), and let . If , then
while if , then by (3.41)
so that Lemma 2.1 implies that
We first show that every solution of (1.2) with has an extension that is a solution of (1.2) such that is a proper subset of . This follows from (c) as soon as it is realized that every solution of (1.2) with has an extension that is a solution of (1.1) such that is a proper subset of . Really, in this case the successive approximations determined by converge to a solution of (1.2), which is obviously an extension of .
That realization can be reached in finitely many steps.
Let be a solution of (1.2) with , and let .

(i)
Suppose .
If , then
is an appropriate solution.
If , then
is a solution of (1.1) that agrees with on .

(ii)
Suppose and . Let
If , then the function
is a solution of (1.1) (, , thus ) that agrees with on .
If , then we introduce the functions , ,
which all lie in , and they are all solutions of (1.1). By Theorem 1.4 (b), is a majorized subset of . Since this space is order complete (see [12]),
exists. Choose . Then almost everywhere on for every , and therefore Lemma 2.1 yields that
Since the proof of Lemma 2.5 shows that , and hence by (A), the function
belongs to too. It now follows from and (3.54) that
We set
By (3.56)
Fix . From Lemma 2.9 (with and there being and resp.) we get that almost everywhere on , and hence
Consequently, , . If with , then . We can see that
By what we have already proved that
is a solution of (1.1) that agrees with on .

(iii)
Suppose and . By an argument entirely similar to that for the case (ii), we can get a solution of (1.1) that agrees with on .
After these preparations, the proof can be concluded quickly. Let be a solution of (1.2), and let be the set of all solutions of (1.2) which agree with on . Since , is not empty. Partially order by declaring to mean that the restriction of to the domain of agrees with . By Hausdorff's maximality theorem, there exists a maximal totally ordered subcollection of . Let be the union of the domains of all members of , and define by , where occurs in . It is easy to check that is well defined, and it is a solution of (1.2). If were a proper subset of , then the first part of the proof would give a further extension of , and this would contradict the maximality of .

(e)
Let be a solution of (1.2), and let .
If , then , so that is bounded on .
Suppose . Since and are bounded on and is finite, Theorem 1.4 (a) implies that it is enough to prove the boundedness of the function
on . Moreover, we have only to observe that the function
is bounded on . To prove this, let
If and , , then and for every , and therefore by Lemma 2.3(d), the function (3.63) is bounded on . If , then the definition of the function gives that . The claim about is therewith confirmed.
The proof of the theorem is now complete.