Answer to Question #170283 in Real Analysis for Prathibha Rose

Question

Suppose alpha increases on [a,b] a less than or equal to x₀ less than or equal to b ,alpha is continuous at x₀ ,f(x₀) =1 and f(x )=0 if x not equal to x₀ .prove that f element. Of R(alpha) and that integral a to b fdalpha =0

Answer 1

Let P = {y0, y1, ..., yn} be a partition of [a, b], i.e

a = y0 ≤ y1 ≤ · · · ≤ ynn1 ≤ yn = b.

We have that there exist yj , j = 0, ..., n, such that x0 = yj or yj-1 < x < yj

. In either case,

we see that

U(P, f, α) ≤ α(yj ) ) α(yj-1) + α(yj+1) ) α(yj ) = α(yj+1) ) α(yj-1),

L(P, f, α) = 0.

Since α is continuous at x0 we see that

|α(yj+1) ) α(yj-1)| → 0 as i → ∞.

So

inf U(P, f, α) = 0.

Therefore f ∈ R(α) and that $\int$f dα = 0.