# The Unapologetic Mathematician

## The Integral as a Function of the Interval

Let’s say we take an integrator $\alpha$ of bounded variation on an interval $\left[a,b\right]$ and a function $f$ that’s Riemann-Stieltjes integrable with respect to $\alpha$ over that interval. Then we know that $f$ is also integrable with respect to $\alpha$ over the subinterval $\left[a,x\right]\subseteq\left[a,b\right]$. Let’s use this to define a function $F$ on $\left[a,b\right]$ by

$\displaystyle F(x)=\int\limits_{\left[a,x\right]}fd\alpha$

We can immediately say some interesting things about this function. First of all, $F$ is, like $\alpha$, of bounded variation. Next, wherever $\alpha$ is continuous, so is $F$. Finally, if $\alpha$ is increasing, then $F$ is differentiable wherever $\alpha$ is differentiable and $f$ is continuous. At such points, we have $F'(x)=f(x)\alpha'(x)$. Notice that, as usual, the first two results will hold if we show that they hold for increasing integrators.

These results are similar to those we get from the Fundamental Theorem of 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 ${x}$ and $y$ in $\left[a,b\right]$, this tells us that

$\displaystyle F(y)-F(x)=\int\limits_{\left[a,y\right]}fd\alpha-\int\limits_{\left[a,x\right]}fd\alpha=\int\limits_{\left[x,y\right]}fd\alpha=f(x_0)\left(\alpha(y)-\alpha(x)\right)$

where $x_0$ is some point between $x$ and $y$.

Now we let $M$ be the supremum of $|F|$ on $\left[a,b\right]$. For any partition of $\left[a,b\right]$ we calculate the variation

$\displaystyle\sum\limits_{i=1}^n|F(x_i)-F(x_{i-1})|\leq M\sum\limits_{i=1}^n|\alpha(x_i)-\alpha(x_{i-1})|\leq MV_\alpha(a,b)$

thus giving an upper bound to the variation of $F$.

Similarly, let ${x}$ be a point where $\alpha$ is continuous. Then given an $\epsilon>0$ we can find a $\delta$ so that $|y-x|<\delta$ implies that $|\alpha(y)-\alpha(x)|<\frac{\epsilon}{M}$. Thus $|F(y)-F(x)|<\epsilon$, and we see that $F$ is continuous at ${x}$ as well.

Finally, we can divide by $y-x$ to find

$\displaystyle\frac{F(y)-F(x)}{y-x}=f(x_0)\frac{\alpha(y)-\alpha(x)}{y-x}$

Then as $y$ gets closer to ${x}$, $x_0$ gets squeezed in towards ${x}$ as well. If $f$ is continuous at ${x}$ and $\alpha$ is differentiable there, then the limit of this difference quotient exists, and has the value stated.

March 31, 2008 - Posted by | Analysis, Calculus

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