Mean Value Theorem for Riemann-Stieltjes Integration

Question

In class we discussed the next theorem:

Let $f$ be a bounded function on $[a,b]$ and $\alpha$ a strictly increasing function on $[a,b]$. Let $f$ be Riemann-Stieltjes integrable with respect to $\alpha$ on [a,b]. If $M=\sup\lbrace f(x):x\in \left[ a,b \right] \rbrace$ and $m=\inf\lbrace f(x):x\in \left[ a,b \right]$

Then $\exists$ $\xi \in \left[ a,b \right]$ such that:

$$ \int_a^b f(x)d\alpha=f(\xi)\left[ \alpha(b) - \alpha(a) \right]$$

We even proved it using partitions, min-max-theorem and intermidiate value theorem.

My question is: Is there a way to prove that $\xi$ can be found in the open interval $\left( a,b \right)$? Have been trying to prove it but I'm kind of lost.. any ideas?

Answer 1

Without more information about $f$, other than bounded and integrable, we can't even be sure that $\xi$ exists.

However, as you may have seen, if $f$ has the intermediate value property (IVP), then there exists at least one $\xi \in [a,b]$ (and possibly more) such that

$$\tag{*} \int_a^b f \, d\alpha=f(\xi)[\alpha(b) - \alpha(a)]$$

For example, $f$ has the IVP if continuous or if (by Darboux's theorem) there is an antiderivative, $f(x) = F'(x)$.

Must there exist $\xi \in (a,b)$, in the open interval, such that (*) holds?

The answer is affirmative.

Defining $m= \inf_{x\in[a,b]} \, f(x)$ and $M = \sup_{x\in[a,b]} \, f(x) $, we know that

$$m \leqslant k =\frac{\int_a^b f \, d\alpha}{\alpha(b) - \alpha(a)} \leqslant M.$$

First, suppose we have strict inequality, $m < k < M.$ By the properties of $\inf$ and $\sup$, there exist $c,d \in [a,b]$ such that $m < f(c) < k < f(d) < M$. Since $f$ has the IVP, there exists $\xi \in (c,d) \subset [a,b]$ such that $f(\xi) = k$. Note that $\xi$ is in the open interval $(c,d)$ since $f(\xi) \neq f(c),f(d)$.

On the other hand, suppose that $k = m$. (The case where $k = M$ is handled in a similar way).

Since $f(x) \geqslant m$ and $\alpha$ is increasing, we have

$$\int_a^b |f(x) - m| \, d\alpha = \int_a^b (f(x) - m) \, d\alpha = (k -m)[\alpha(b) - \alpha(a)] = 0,$$

and it follows that $f(x) = m$ almost everywhere and there exists $\xi \in (a,b)$ such that $f(\xi) = m = k$.