an elementary proof of lebesgue's differentiation theorem

3 MAIN RESULTNow we are going to establish the main result. We will achieve the resultby using the lemmas in preliminary part. We need them in some technicalpart of our proof. Our proof will depend on no measure theory beyond setsof measure zero. Also upper and lower derivatives are used.Theorem: If f is nondecreasing on [a, b], then f ′ (x) exists almost everywhereon [a, b].Proof: We know that the only discontinuities of monotonic functions arejumps [2, p. 129]. Since f is continuous except at a countable number ofpoints, it is sufficient to show thatF = {x : x ∈ (a, b), f is continuous at x and Df(x) > Df(x)}has measure zero. HereDf(x) = lim supy→xf(y) − f(x)f(y) − f(x), Df(x) = lim infy − xy→x y − xare the upper and lower derivatives of f at x, respectively. Clearly F is thecountable union of the setsE r,s = {x : x ∈ (a, b), f is continuous at x and Df(x) > r > s > Df(x)}for rational numbers r and s with r > s. Thus we need to show that eachE r,s has measure zero.Assume for a contradiction that for some choice of r and s the set E = E r,sdoes not have measure zero, and let ε be the positive number in Lemma 1.Let A = (r − s)/2, B = (r + s)/2, and g = f − Bx.and B are positive andClearly both AE = {x : x ∈ (a, b), g is continuous at x, Dg(x) > A and Dg(x) < −A}.Now { ∑ P |g(x k)−g(x k−1 )|: P is partition of [a, b]} is bounded above. Namely,∑|g(x k ) − g(x k−1 )| = ∑ |f(x k ) − f(x k−1 ) − B(x k − x k−1 )|PP≤ ∑ |f(x k ) − f(x k−1 )| + ∑ B|(x k − x k−1 )|PP= f(b) − f(a) + B(b − a).6