Riemann Integral of Functions from ℝ into Real Banach Space

Summary In this article we deal with the Riemann integral of functions from R into a real Banach space. The last theorem establishes the integrability of continuous functions on the closed interval of reals. To prove the integrability we defined uniform continuity for functions from R into a real normed space, and proved related theorems. We also stated some properties of finite sequences of elements of a real normed space and finite sequences of real numbers. In addition we proved some theorems about the convergence of sequences. We applied definitions introduced in the previous article [21] to the proof of integrability.


Summary.
In this article we deal with the Riemann integral of functions from R into a real Banach space. The last theorem establishes the integrability of continuous functions on the closed interval of reals. To prove the integrability we defined uniform continuity for functions from R into a real normed space, and proved related theorems. We also stated some properties of finite sequences of elements of a real normed space and finite sequences of real numbers.
In addition we proved some theorems about the convergence of sequences. We applied definitions introduced in the previous article [21] to the proof of integrability.

Some Properties of Continuous Functions
In this paper s 1 , s 2 , q 1 denote sequences of real numbers. Let X be a real normed space and f be a partial function from R to the carrier of X. We say that f is uniformly continuous if and only if (Def. 1) Let us consider a real number r. Suppose 0 < r. Then there exists a real number s such that (i) 0 < s, and (ii) for every real numbers Now we state the propositions: (1) Let us consider a set X, a real normed space Y , and a partial function f from R to the carrier of Y . Then f X is uniformly continuous if and only if for every real number r such that 0 < r there exists a real number s such that 0 < s and for every real numbers Proof: If f X is uniformly continuous, then for every real number r such that 0 < r there exists a real number s such that 0 < s and for every real numbers [11, (80)]. Consider s being a real number such that 0 < s and for every real numbers (2) Let us consider sets X, X 1 , a real normed space Y , and a partial function f from R to the carrier of Y . Suppose (i) f X is uniformly continuous, and Then f X 1 is uniformly continuous. The theorem is a consequence of (1).
(3) Let us consider a real normed space X, a partial function f from R to the carrier of X, and a subset Z of R. Suppose Then f Z is uniformly continuous. The theorem is a consequence of (1).

Some Properties of Sequences
Now we state the proposition: (4) Let us consider a real normed space X, natural numbers n, m, a function a from Seg n × Seg m into X, and finite sequences p, q of elements of X. Suppose (ii) for every natural number i such that i ∈ dom p there exists a finite sequence r of elements of X such that dom r = Seg m and p(i) = r and for every natural number j such that j ∈ dom r holds r(j) = a(i, j), and (iii) dom q = Seg m, and (iv) for every natural number j such that j ∈ dom q there exists a finite sequence s of elements of X such that dom s = Seg n and q(j) = s and for every natural number i such that i ∈ dom s holds s(i) = a(i, j).
Then p = q. Proof: Define P[natural number] ≡ for every natural number m for every function a from Seg $ 1 × Seg m into X for every finite sequences p, q of elements of X such that dom p = Seg $ 1 and for every natural number i such that i ∈ dom p there exists a finite sequence r of elements of X such that dom r = Seg m and p(i) = r and for every natural number j such that j ∈ dom r holds r(j) = a(i, j) and dom q = Seg m and for every natural number j such that j ∈ dom q there exists a finite sequence s of elements of X such that dom s = Seg $ 1 and q(j) = s and for every natural number i such that i ∈ dom s holds s(i) = a(i, j) holds p = q. For every natural number n such that P[n] holds P[n + 1] by [4, (5)], [2, (11) Then q = vol(A). Proof: Set p = lower volume( χ A,A , D). For every natural number k such that k ∈ dom q holds q(k) = p(k) by [15, (19)]. (ii) for every natural number i such that i ∈ dom S there exists a real number r such that r = q(i) and S(i) = r · E.
Then S = q · E. Proof: Define P[natural number] ≡ for every finite sequence q of elements of R for every finite sequence S of elements of Y such that $ 1 = len S and len S = len q and for every natural number i such that i ∈ dom S there exists a real number r such that r = q(i) and Then middle sum(h, S) is convergent. The theorem is a consequence of (8), (7), (4), (12), (5), (10), and (6). Proof: For every division sequence T of A and for every middle volume sequence S of h and T such that δ T is convergent and lim δ T = 0 holds middle sum(h, S) is convergent by [32, (57)], [15, (9)], [17, (9)].
The scheme ExRealSeq2X deals with a non empty set D and a unary functor F, G yielding an element of D and states that (Sch. 1) There exists a sequence s of D such that for every natural number n, s(2 · n) = F(n) and s(2 · n + 1) = G(n).
Now we state the propositions: (14) Let us consider a natural number n. Then there exists a natural number k such that n = 2 · k or n = 2 · k + 1.  Then there exists a middle volume sequence S 1 of h and T 1 such that for every natural number i, S 1 (2 · i) = S 7 (i) and S 1 (2 · i + 1) = S(i).
The theorem is a consequence of (14). Proof: Reconsider S 2 = S 7 , S 3 = S as a sequence of (the carrier of X) * . Define F(natural number) = S 2$ 1 . Define G(natural number) = S 3$ 1 . Consider S 1 being a sequence of (the carrier of X) * such that for every natural number n, S 1 (2 · n) = F(n) and S 1 (2 · n + 1) = G(n) from ExRealSeq2X. For every element i of N, S 1 (i) is a middle volume of h and T 1 (i). The theorem is a consequence of (14). Proof: For every real number r such that 0 < r there exists a natural number m 1 such that for every natural number i such that m 1 i holds S 4 (i) − lim S 5 < r by [2, (11)]. For every real number r such that 0 < r there exists a natural number m 1 such that for every natural number i such that m 1 i holds S 6 (i) − lim S 5 < r by [2, (11)]. Reconsider h = f A as a function from A into the carrier of X. Consider T 2 being a division sequence of A such that δ T 2 is convergent and lim δ T 2 = 0. Set S 7 = the middle volume sequence of h and T 2 . Set I = lim middle sum(h, S 7 ). For every division sequence T of A and for every middle volume sequence S of h and T such that δ T is convergent and lim δ T = 0 holds middle sum(h, S) is convergent and lim middle sum(h, S) = I.