The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


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


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.

December 29, 2009 - Posted by | Analysis, Calculus


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