# The Second Mean Value Theorem for Integrals

This post concerns the second mean value theorem for definite integrals.

Theorem    [Second MVT for Integrals]
Let $w$ be a bounded, increasing real valued function defined on the interval $[a,b]$, that is continuous at $a$ and $b$; and $f$ a Riemann integrable function on $[a,b]$. Then there exists $c\in [a,b]$ such that $w(a) \int_a^c f(x)~dx + w(b) \int_c^b f(x)~dx = \int_a^b w(x) f(x)~dx.$

The proof of this theorem is actually similar to the proof of the integration by parts formula for Riemann integrable functions.

First notice that since $w$ is monotone, its set of discontinuity points is countable and hence has measure zero. By Lebesgue's criterion $w$ is integrable, and so the integral $\int_a^b w(x) f(x)~dx$ is well-defined.

Let $F(x) = \int_a^x f(s)~dx$. The LHS can be written as $w(b) F(b) - w(a) F(a) - (w(b) - w(a)) F(c)$ and we will try to prove the existence of $c$ such that $w(b) F(b) - w(a) F(a) - \int_a^b w(x) f(x)~dx = (w(b) - w(a)) F(c).$ Since we know that $F$ is a continuous function, it suffices by the intermediate value theorem to prove that $[ w(b) - w(a)] \inf F([a,b]) \leq w(b) F(b) - w(a) F(a) - \int_a^b w(x) f(x) ~dx \leq [w(b) - w(a)] \sup F([a,b]).$

Let $\mathcal{D}$ denote a division of $[a,b]$ into finitely many closed intervals; in other words, elements of $\mathcal{D}$ are closed subintervals of $[a,b]$ and there exists a way to list the elements so that $\sup I_i = \inf I_{i+1}$ for intervals $I_i$ and $I_{i+1}$ in $\mathcal{D}$.

We can write $w(b) F(b) - w(a) F(a) = \sum_{I\in \mathcal{D}} w(\sup I) F(\sup I) - w(\inf I) F(\inf I) = \sum_{I\in \mathcal{D}} [w(\sup I) - w(\inf I) ] F(\sup I) + w(\inf I) [ F(\sup I) - F(\inf I)]$ Since $w$ is increasing, we have $w(\sup I) - w(\inf I) > 0$. And therefore we clearly have that the first sum $[ w(b) - w(a)] \inf F([a,b]) \leq \sum_{I \in \mathcal{D}} [w (\sup I) - w(\inf I)] F(\sup I) \leq [ w(b) - w(a)] \sup F([a,b]).$ Thus it remains to prove that $\sum_{I\in \mathcal{D}} w(\inf I) [ F(\sup I) - F(\inf I)] - \int_a^b w(x) f(x)~dx$ can be made arbitrarily small. This sum can be rewritten as $\sum_{I \in \mathcal{D}} \int_I (w(\inf I) - w(x)) f(x)~dx.$ If $w$ were continuous, then by uniform continuity, for every $\epsilon > 0$, we can pick $\delta > 0$ so that if $\mathcal{D}$ is $\delta$-fine, then this sum can be bounded by $\epsilon \int_a^b |f(x)|~dx$, and the theorem follows.

For general increasing $w$, however, what we can do is use that $w$ is Riemann integrable. Let $M > \sup |f([a,b])|$. Let $X\subseteq [a,b]$ denote the finite set of points at which the oscillation of $w$ is greater than $\epsilon'$, for some $\epsilon'$ to be fixed. (This set is finite since by monotonicity, the total change $w(b) - w(a) \geq \sum \epsilon' |X|$.) Choose our $\mathcal{D}$ such that it is $\delta$ fine for $\delta < \frac{\epsilon}{2|X||b-a| M}$. Then we have $\Big|\sum_{I \in \mathcal{D}\land I\cap X\neq\emptyset} \int_I (w(\inf I) - w(x)) f(x)~dx\Big| \leq |X| \delta |b-a| M < \frac{\epsilon}{2}.$ Here we used that since $w$ is monotone, its oscillation on any interval is no more than $b-a$.

Next, $[a,b]\setminus X$ is compact. Therefore there exists $r:[a,b]\setminus X$ such that for every $x\in [a,b]\setminus X$, the oscillation of $w$ on $(x-2r(x), x+2r(x))$ is no more than $2\epsilon'$. The set $\{ (x - r(x), x + r(x)) : x\in [a,b]\setminus X\}$ forms an open cover, and hence has a finite subcover. Let the $r$ denote the minimum radius of the finite subcover.

Now if we make sure $\delta < r$, this means that any $I\in \mathcal{D}$ that does not intersect $X$ must be contained in $(x-2r(x), x+2r(x))$ for some $x$ indexing the finite subcover. And this implies that the oscillation of $w$ on $I$ is no more than $2\epsilon'$. So $\Big|\sum_{I \in \mathcal{D}\land I\cap X = \emptyset} \int_I (w(\inf I) - w(x)) f(x)~dx\Big| \leq 2 \epsilon' \int_a^b |f(x)|~dx.$ And thus if we had chosen $\epsilon' < \frac{1}{4\epsilon \int_a^b |f(x)| ~dx}.$ then this sum would also contribute no more than $\epsilon /2$.

Since $\epsilon$ is arbitrary, we have proven the theorem.

Notice that if $\int_a^b |f(x)| ~dx = 0$ then $f$ vanishes almost everywhere and the theorem holds trivially; so we can rule out this case when defining $\epsilon'$. ##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.