實變函數論

Question

If f is absoutely continuous on [a,b],then it is of bounded variation on [a,b]Prove it

Answer 1

Proof. By the absolute continuity of f, there exists δ > 0 such that for any finite collection of disjoint intervals (ai,bi) ⊆ [a,b], i = 1,2,...,M, with∑i=1M (bi - ai) < δ.we have∑i=1M |f(bi) - f(ai)| < 1.Choose N ∈ N large enough so that (b-a)/N < δ. I will show that the total variation of f is at most N.Given a partitionP: a = x0 < x1 < ... < xn = b,we can create a further partitionP': a = y0 < y1 < ... < ym = b,by adding points a + i(b-a)/N for i = 1,2,...,N-1 to P.Now suppose j(i) is the index that has yj(i) = a + i(b-a)/N for i ∈ {0,1,...,N}.For each i ∈ {0,1,...,N-1},∑j=j(i)+1j(i+1) (yj - yj-1)= yj(i+1) - yj(i) = (a + (i+1)(b-a)/N) - (a + i(b-a)/N)= (b-a)/N < δ,hence∑j=j(i)+1j(i+1) |f(yj) - f(yj-1)| < 1,and so∑j=1n |f(xj) - f(xj-1)| ≤ ∑j=1m |f(yj) - f(yj-1)| = ∑i=0N-1 ∑j=j(i)+1j(i+1) |f(yj) - f(yj-1)| < N.Therefore, f is of bounded variation. ∎