# The Unapologetic Mathematician

## Two Mean Value Theorems

We’ve got two different analogues of the integral mean value theorem for the Riemann-Stieltjes integral.

The first one says that if $\alpha$ is increasing on $\left[a,b\right]$ and $f$ is integrable with respect to $\alpha$, with supremum $M$ and infimum $m$ in the interval, then there is some “average value” $c$ between $m$ and $M$. This satisfies

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=c\int\limits_{\left[a,b\right]}d\alpha=c\left(\alpha(b)-\alpha(a)\right)$

In particular, we should note that if $f$ is continuous then the intermediate value theorem tells us that there is some $x_0$ with $f(x_0)=c$. That is, there is some $x_0$ such that

$\displaystyle f(x_0)=\frac{1}{\alpha(b)-\alpha(a)}\int\limits_{\left[a,b\right]}fd\alpha$

When $\alpha(x)=x$ this gives us the old integral mean value theorem back again.

So why does this work? Well, if $\alpha(a)=\alpha(b)$ then both sides are zero and the theorem is trivially true. Now, the lowest lower sum is $L_{\alpha,\{a,b\}}(f)=m\left(\alpha(b)-\alpha(a)\right)$, while the highest upper sum is $U_{\alpha,\{a,b\}}(f)=M\left(\alpha(b)-\alpha(a)\right)$. The integral itself, which we’re assuming to exist, lies between these bounds:

$\displaystyle m\left(\alpha(b)-\alpha(a)\right)\leq\int\limits_{\left[a,b\right]}fd\alpha\leq M\left(\alpha(b)-\alpha(a)\right)$

So we can divide through by $\int_{\left[a,b\right]}d\alpha=\alpha(b)-\alpha(a)$ to get the result we seek.

We can get a similar result which focuses on the integrator by using integration by parts. Let’s assume $\alpha$ is continuous and $f$ is increasing on $\left[a,b\right]$. Our sufficient conditions tell us that the integral of $f$ with respect to $\alpha$ exists, and the integration by parts formula says

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\int\limits_{\left[a,b\right]}\alpha df$

But the first integral mean value theorem tells us that the integral on the right is equal to $\alpha(x_0)\left(f(b)-f(a)\right)$ for some $x_0$. Then we can rearrange the above formula to read

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\alpha(x_0)\left(f(b)-f(a)\right)$
$\displaystyle=f(a)\left(\alpha(x_0)-\alpha(a)\right)+f(b)\left(\alpha(b)-\alpha(x_0)\right)$
$\displaystyle=f(a)\int_{\left[a,x_0\right]}d\alpha+f(b)\int_{\left[x_0,b\right]}d\alpha$

So there is some point $x_0$ so that the integral of $f$ is the same as the integral of the step function taking the value $f(a)$ until $x_0$ and the value $f(b)$ after it.

March 28, 2008 - Posted by | Analysis, Calculus

## 4 Comments »

1. […] under the same conditions as the integral version. And I’ll do this by using a form of the Integral Mean Value Theorem for Riemann-Stieltjes […]

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2. […] Calculus, and we can use some of the same techniques to prove them. In particular, we call on the Integral Mean-Value Theorem for Riemann-Stieltjes integrals. If we take points and in , this tells us […]

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3. […] giving bounds on the integral in terms of the Jordan content of . Incidentally, here is serving a similar role to the integrator in the integral mean value theorem for Riemann-Stieltjes integrals. […]

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4. […] we’ve used the integral mean value theorem. Clearly by choosing the right we can find a to make the right hand side as small as we want, […]

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