The Unapologetic Mathematician

Mathematics for the interested outsider

The Riemann-Stieltjes Integral I

Today I want to give a modification of the Riemann integral which helps give insight into the change of variables formula.

So, we defined the Riemann integral

\displaystyle\int\limits_a^bf(x)dx

to be the limit as we refined the tagged partition x=((x_0,...,x_n),(t_1,...,t_n)) of the Riemann sum

\displaystyle f_x=\sum\limits_{i=1}^nf(t_i)(x_i-x_{i-1})

But why did we multiply by (x_i-x_{i-1})? Well, that was the width of a rectangular strip we were using to approximate part of the area under the graph of f. But why should we automatically use that difference as the “width”?

Let’s imagine we’re walking past a fence. Sometimes we walk faster, and sometimes we walk slower, but at time t we can measure the height of the fence right next to us: f(t). So what’s the area of the fence? If we just integrated f(t) we’d get the wrong answer. The samples we made when walking fast made fat rectangles, while the samples we made when we were walking slowly got paired with skinny rectangles, but we gave them the same weight if they took the same time to get through that segment of the partition. We need to reweight our sums to compensate for how fast we’re walking!

Okay, so how wide should we make the rectangles? Let’s say that at time t we’re at position \alpha(t) along the fence. Then in the segment of the partition between times x_{i-1} and x_i we move from position \alpha(x_{i-1}) to position \alpha(x_i), so we should make the width come out to (\alpha(x_i)-\alpha(x_{i-1})). We’ll put this into our formalism from before and get the “Riemann-Stieltjes sum”:

\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)(\alpha(x_i)-\alpha(x_{i-1}))

And now we can take the limit over tagged partitions as before to get the “Riemann-Stieltjes integral”:

\displaystyle\int\limits_{\left[a,b\right]}f(x)d\alpha(x)=\int\limits_a^bf(x)d\alpha(x)

if this limit exists.

Here we call the function f the “integrand”, and the function \alpha the “integrator”. Clearly, the old Riemann integral is the special case when \alpha(x)=x.

Immediately from the definition we can see the same “additivity” (using signed intervals) in the region of integration that the Riemann integral had:

\displaystyle\int\limits_{\left[x_1,x_3\right]+\left[x_3,x_2\right]}f(x)d\alpha(x)=\int\limits_{\left[x_1,x_3\right]}f(x)d\alpha(x)+\int\limits_{\left[x_3,x_2\right]}f(x)d\alpha(x)

and the same linearity in the integrand:

\displaystyle\int\limits_{\left[x_1,x_2\right]}af(x)+bg(x)d\alpha(x)=a\int\limits_{\left[x_1,x_2\right]}f(x)d\alpha(x)+b\int\limits_{\left[x_1,x_2\right]}g(x)d\alpha(x)

and also a new linearity in the integrator:

\displaystyle\int\limits_{\left[x_1,x_2\right]}f(x)d(a\alpha(x)+b\beta(x))=a\int\limits_{\left[x_1,x_2\right]}f(x)d\alpha(x)+b\int\limits_{\left[x_1,x_2\right]}f(x)d\beta(x)

Neat!

February 28, 2008 Posted by John Armstrong | Analysis, Calculus | | 15 Comments

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The Big Word Project has the answer.

February 28, 2008 Posted by John Armstrong | Uncategorized | | 3 Comments