I have just been reading about the mean value theorems for integrals, surface integrals and line integrals. I did a Google search for a corresponding theorem of volume integrals, and couldn't find any evidence of one.

I would guess that, if $f(x,y,z)$ is a differentiable function defined in the region $\Omega$ in $\mathbb{R}^3$, then:


where $(x_0,y_0,z_0)$ is some point in $\Omega$ and $V(\Omega)$ is the volume of the region $\Omega$.

This holds for the trivial example $f(x,y,z)=c$, where $c$ is a constant in the domain of interest. Are there counter-examples, or does such a theorem exist?

  • $\begingroup$ Pretty straightforward. $\endgroup$ – A.Γ. Nov 19 '16 at 21:36
  • $\begingroup$ Does this need differentiability, or just continuity? $\endgroup$ – Teepeemm Nov 20 '16 at 0:45

Since $f$ is differentiable, it is continuous. We assume first that $\Omega$ is a compact connected set (any such set is automatically Lebesgue-measurable).

We have $m=\inf f \leq f \leq \sup f=M$, and

$$m\mu(\Omega) \leq \int_{\Omega} f \leq M \mu(\Omega), $$

(note that $\mu(\Omega)<\infty$), from which

$$ m \leq \frac{\int_{\Omega} f}{\mu(\Omega)} \leq M,$$

and the result follows from the "intermediate value theorem" (or rather the fact that continuous maps preserve connected sets). (Note also that the case $\mu(\Omega)=0$ is trivial).

One can adapt the proof above to a more general case: that of $\Omega$ connected, Lebesgue-measurable and with finite measure, and $f$ continuous and integrable (which I think is the most general we can have, since dropping any of the hypotheses lead to counterexamples or simply the statement not making sense). Firstly, note that if $f$ is not bounded, the result is trivial. Now, note that we have in the same way as above: $$m\mu(\Omega) \leq \int_{\Omega} f \leq M \mu(\Omega). $$ Now, if $m\mu(\Omega)=\int _{\Omega} f,$ we have $\int_{\Omega} f-m=0$, and it follows that $f=\inf f$ almost everywhere (since $f-m \geq 0$). In particular, $f$ assumes its infinum somewhere (we are assuming $\mu(\Omega) \neq 0$. If not, then the result is already trivial from the start). Analogously, we have that if $M \mu(\Omega)=\int_{\Omega}f$, then $f$ assumes its supremum somewhere.

This is just to say that, in any case, we have $\frac{\int_{\Omega} f}{\mu(\Omega)}$ belonging to the image of $f$.

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    $\begingroup$ You might consider showing that the IVT applies here by using a single parameter description of a curve in the domain that connects the points at which $f$ attains its min and max. $\endgroup$ – Mark Viola Nov 19 '16 at 21:50
  • $\begingroup$ The IVT applies here because $f$ is a continuous real function on a connected set. It is just a matter of names, and I added clarification. $\endgroup$ – Aloizio Macedo Nov 19 '16 at 21:51
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    $\begingroup$ Yes, I know. But you're not answering the question for yourself, you're answering for those who don't see the extension from a single variable to multiple ones. $\endgroup$ – Mark Viola Nov 19 '16 at 21:55
  • $\begingroup$ Thanks for the response. Unfortunately I don't know enough to understand it, as I am a non-expert. Not your fault, obviously, but I am meant to select the answer which I find the most useful. $\endgroup$ – Lachy Nov 19 '16 at 22:17
  • $\begingroup$ @Lachy That is no problem! Your conception of what answer to accept is the intended one. : ) $\endgroup$ – Aloizio Macedo Nov 19 '16 at 22:20

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