We’ve got two different analogues of the integral mean value theorem for the Riemann-Stieltjes integral.
The first one says that if 
is increasing on ![\left[a,b\right]](http://s0.wp.com/latex.php?latex=%5Cleft%5Ba%2Cb%5Cright%5D&bg=e6e6e6&fg=333333&s=0 "\left[a,b\right]")
and 
is integrable with respect to , with supremum 
and infimum 
in the interval, then there is some “average value” 
between and . 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)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%3Dc%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dd%5Calpha%3Dc%5Cleft%28%5Calpha%28b%29-%5Calpha%28a%29%5Cright%29&bg=e6e6e6&fg=333333&s=0 "\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 is continuous then the intermediate value theorem tells us that there is some 
with 
. That is, there is some such that
![\displaystyle f(x_0)=\frac{1}{\alpha(b)-\alpha(a)}\int\limits_{\left[a,b\right]}fd\alpha](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+f%28x_0%29%3D%5Cfrac%7B1%7D%7B%5Calpha%28b%29-%5Calpha%28a%29%7D%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha&bg=e6e6e6&fg=333333&s=0 "\displaystyle f(x_0)=\frac{1}{\alpha(b)-\alpha(a)}\int\limits_{\left[a,b\right]}fd\alpha")
When 
this gives us the old integral mean value theorem back again.
So why does this work? Well, if 
then both sides are zero and the theorem is trivially true. Now, the lowest lower sum is 
, while the highest upper sum is 
. 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)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+m%5Cleft%28%5Calpha%28b%29-%5Calpha%28a%29%5Cright%29%5Cleq%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%5Cleq+M%5Cleft%28%5Calpha%28b%29-%5Calpha%28a%29%5Cright%29&bg=e6e6e6&fg=333333&s=0 "\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)](http://s0.wp.com/latex.php?latex=%5Cint_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dd%5Calpha%3D%5Calpha%28b%29-%5Calpha%28a%29&bg=e6e6e6&fg=333333&s=0 "\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 is continuous and is increasing on . Our sufficient conditions tell us that the integral of with respect to 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](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%3Df%28b%29%5Calpha%28b%29-f%28a%29%5Calpha%28a%29-%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7D%5Calpha+df&bg=e6e6e6&fg=333333&s=0 "\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 
for some . 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)](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint%5Climits_%7B%5Cleft%5Ba%2Cb%5Cright%5D%7Dfd%5Calpha%3Df%28b%29%5Calpha%28b%29-f%28a%29%5Calpha%28a%29-%5Calpha%28x_0%29%5Cleft%28f%28b%29-f%28a%29%5Cright%29&bg=e6e6e6&fg=333333&s=0 "\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)\int_{\left[a,x_0\right]}d\alpha+f(b)\int_{\left[x_0,b\right]}d\alpha](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%3Df%28a%29%5Cint_%7B%5Cleft%5Ba%2Cx_0%5Cright%5D%7Dd%5Calpha%2Bf%28b%29%5Cint_%7B%5Cleft%5Bx_0%2Cb%5Cright%5D%7Dd%5Calpha&bg=e6e6e6&fg=333333&s=0 "\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 so that the integral of is the same as the integral of the step function taking the value 
until and the value 
after it.
March 28, 2008
Posted by John Armstrong |
Analysis, Calculus |
4 Comments
