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Let $α ∈ C([a, b]) ∩ BV([a, b])$. Assume $g ∈ R(α)$ on $[a, b]$ and define $β(x) = \int_a^xg(t)dα(t)$ if $x ∈ [a, b]$. Show that

(a) if $f$ is increasing on $[a, b]$, there exists $x0 ∈ [a, b]$ such that

$$\int_a^b f~d\beta=f(a)\int_x^{x_0}g~d\alpha+f(b)\int_{x_0}^bg~d\alpha,$$

(b) if, in addition, $f$ is continuous on $[a,b],$ we also have

$$\int_a^b f(x)g(x)~d\alpha(x)=f(a)\int_a^{x_0}g~d\alpha+f(b)\int_{x_0}^bg ~d\alpha,$$

I have proved the first problem using the integration by parts formula and the first mean value theorem. However, I got stuck on the second problem. We don't know anything about $\alpha'(x)$, so we cannot use differential formula. I also tried to define a function $F(x)=f(a)\int_a^{x}g~d\alpha+f(b)\int_{x}^bg ~d\alpha$ to prove that $\int_a^b f(x)g(x)~d\alpha(x)$ lies between $F(a)$ and $F(b)$, but it failed because $\int_a^b f(x)g(x)~d\alpha(x)$ lies between $f(a)\int_a^bg~d\alpha$ and $f(b)\int_a^bg~d\alpha$ is not true.

I have tried to use the definition to prove it. It still did not work.

I totally have no idea about the second problem right now.

  • 1
    From part (a), you know that $$f(a) \int_a^{x_0} g d\alpha + f(b) \int_{x_0}^b g d\alpha = \int_a^b fd\beta$$ and $$f(a) \int_a^{x_0} g d\alpha + f(b) \int_{x_0}^b g d\alpha = \int_a^b f(x)g(x)d\alpha(x)$$ so can you show that $d\beta = g(x)d\alpha(x)$? – 2017-02-02
  • 0
    @homedoggieo I don't know how to prove it because the derivative of $/alpha(x)$ may not exist. – 2017-02-02

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