# The Unapologetic Mathematician

## Continuity of Partial Integrals

There are some remaining topics to clean up in the theory of the Riemann-Stieltjes integral. First up is a question that seems natural from the perspective of iterated integrals: what can be said about the continuity of the inner integrals?

More explicitly, let $f:R\rightarrow\mathbb{R}$ be a continuous function on the rectangle $R=[a,b]\times[c,d]$, and let $\alpha$ be of bounded variation on $[a,b]$. Define the function $F$ on $[c,d]$ by

$\displaystyle F(y)=\int\limits_a^bf(x,y)\,d\alpha(x)$

Then $F$ is continuous on $[c,d]$. Similar statements can be made about other “partial integrals”, where $x$ and $y$ are each vector variables and we use $d(x^1,\dots,x^k)$ in place of the Stieltjes integrator $d\alpha(x)$.

Specifically, this is a statement about interchanging limit operations. The Riemann-Stieltjes integral involves taking the limit over the collection of tagged partitions of $[a,b]$, while to ask if $F$ is continuous asks whether

$\displaystyle\lim\limits_{y\rightarrow y_0}\int\limits_a^bf(x,y)\,d\alpha(x)=\int\limits_a^b\lim\limits_{y\rightarrow y_0}f(x,y)\,d\alpha(x)=\int\limits_a^bf(x,y_0)\,d\alpha(x)$

As we knew back when we originally discussed integrators of bounded variation, we can write our integrator as the difference of two increasing functions. It’s no loss of generality, then, to assume that $\alpha$ is increasing. We also remember that the Heine-Cantor theorem tells us that since $R$ is compact, $f$ is actually uniformly continuous.

Uniform continuity tells us that for every $\epsilon>0$ there is a $\delta>0$ (depending only on $\epsilon$ so that for every pair of points $(x,y)$ and $(x',y')$, with $\lvert(x,y)-(x',y')\rvert<\delta$ we have $\lvert f(x,y)-f(x',y')\rvert<\epsilon$.

So now let’s take two points $y$ and $y'$ with $\lvert y-y'\rvert<\delta$ and consider the difference

\displaystyle\begin{aligned}\lvert F(y)-F(y')\rvert&=\left\lvert\int\limits_a^bf(x,y)\,d\alpha(x)-\int\limits_a^bf(x,y')\,d\alpha(x)\right\rvert\\&=\left\lvert\int\limits_a^bf(x,y)-f(x,y')\,d\alpha(x)\right\rvert\\&\leq\int\limits_a^b\lvert f(x,y)-f(x,y')\rvert\,d\alpha(x)\\&\leq\epsilon\left(\alpha(b)-\alpha(a)\right)\end{aligned}

where we’ve used the integral mean value theorem. Clearly by choosing the right $\epsilon$ we can find a $\delta$ to make the right hand side as small as we want, proving the continuity of $F$.