# The Unapologetic Mathematician

## The Mean Value Theorem for Multiple Integrals

As in the single variable case, multiple integrals satisfy a mean value property.

First of all, we should note that, like one-dimensional Riemann-Stieltjes integrals with increasing integrators, integration preserves order. That is, if $f$ and $g$ are both integrable over a Jordan-measurable set $S$, and if $f(x)\leq g(x)$ at each point $x\in S$, then we have

$\displaystyle\int\limits_Sf(x)\,dx\leq\int\limits_Sg(x)\,dx$

This is a simple consequence of the definition of a multiple integral as the limit of Riemann sums, since every Riemann sum for $f$ will be smaller than the corresponding sum for $g$.

Now if $f$ and $g$ are integrable on $S$ and $g(x)\geq0$ for every $x\in S$, then we set $m=\inf f(S)$ and $M=\sup f(S)$ — the infimum and supremum of the values attained by $f$ on $S$. I assert that there is some $\lambda$ in the interval $[m,M]$ so that

$\displaystyle\int\limits_Sf(x)g(x)\,dx=\lambda\int\limits_Sg(x)\,dx$

In particular, we can set $g(x)=1$ and find

$\displaystyle mc(S)\leq\int\limits_Sf(x)\,dx\leq Mc(S)$

giving bounds on the integral in terms of the Jordan content of $S$. Incidentally, $g(x)\,dx$ here is serving a similar role to the integrator $d\alpha$ in the integral mean value theorem for Riemann-Stieltjes integrals.

Okay, so since $g(x)\geq0$ we have $mg(x)\leq f(x)g(x)\leq Mg(x)$ for every $x\in S$. Since integration preserves order, this yields

$\displaystyle m\int\limits_Sg(x)\,dx\leq\int\limits_Sf(x)g(x)\,dx\leq M\int\limits_Sg(x)\,dx$

If the integral of $g$ is zero, then our result automatically holds for any value of $\lambda$. Otherwise we can divide through by this integral and set

$\displaystyle\lambda=\frac{\displaystyle\int\limits_Sf(x)g(x)\,dx}{\displaystyle\int\limits_Sg(x)\,dx}$

which will be between $m$ and $M$.

One particularly useful case is when $S$ has Jordan content zero. In this case, we find that any integral over $S$ is itself automatically zero.

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Posted by John Armstrong | Analysis, Calculus

## 3 Comments	»

1. [...] in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no [...]

Pingback by | December 30, 2009 | Reply

2. [...] infinitesimal pieces of -dimensional volume. Now, with the change of variables formula and the mean value theorem in hand, we can pull out a macroscopic [...]

Pingback by | January 8, 2010 | Reply

3. [...] analogue of the integral mean value theorem that holds not just for single integrals, not just for multiple integrals, but for integrals over any measure [...]

Pingback by | June 14, 2010 | Reply

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