Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

==== If you insist on defining the definite integral solely via values of antiderivatives and your extended linear limit LLL, you must: ====
# Define continuity of f:[0,t]→Ef:[0,t]\to Ef:[0,t]→E in terms of the limit LLL: i.e. fff is continuous at s0s_0s0 if for every sequence sn→s0s_n\to s_0sn→s0 (classically) we have Ln→∞f(sn)=f(s0)L_{n\to\infty} f(s_n)=f(s_0)Ln→∞f(sn)=f(s0). But that presupposes a way to take limits of sequences with values in EEE — you must make that precise.
# Construct a primitive F(x):=Ln→∞∑k=1nf(ξk)ΔxkF(x) := L_{n\to\infty} \sum_{k=1}^{n} f(\xi_k)\Delta x_kF(x):=Ln→∞∑k=1nf(ξk)Δxk (Riemann sums) and prove F′(x)=f(x)F'(x)=f(x)F′(x)=f(x) in the sense of your extended derivative. This again requires that LLL commutes with the relevant operations and that uniform continuity estimates hold to pass the limiting operation through subtraction/division.
# This route is essentially reproducing the classical proof of the FTC but with LLL replacing the classical limit. For it to succeed you need stronger and more explicit axioms: LLL must preserve uniform convergence and commute with multiplication by scalar step functions, etc. These are nontrivial axioms and are exactly the hidden extra assumptions you were trying to avoid.

Why this route is brittle: you again need control of sequences of Riemann sums and uniformity; the “antiderivative exists because the integrand is continuous” claim becomes equivalent to proving that your extension of limits has the same uniform-convergence properties as the classical limit — a major extra hypothesis.