# The Unapologetic Mathematician

## The Integral Mean Value Theorem

We have an 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 space.

If $f$ is an essentially bounded measurable function with $\alpha\leq f\leq\beta$ a.e. for some real numbers $\alpha$ and $\beta$, and if $g$ is any integrable function, then there is some real number $\gamma$ with $\alpha\leq\gamma\leq\beta$ so that

$\displaystyle\int f\lvert g\rvert\,d\mu=\gamma\int\lvert g\rvert\,d\mu$

Actually, this is a statement about finite measure spaces; the function $g$ is here so that the indefinite integral of $\lvert g\rvert$ will give us a finite measure on the measurable space $X$ to replace the (possibly non-finite) measure $\mu$. This explains the $g$ in the multivariable case, which wasn’t necessary when we were just integrating over a finite interval in the one-variable case.

Okay, we know that $\alpha\leq f\leq\beta$ a.e., and so $\alpha\lvert g\rvert\leq f\lvert g\rvert\leq\beta\lvert g\rvert$ a.e. as well. This tells us that $f\lvert g\rvert$ is integrable. And thus we conclude

$\displaystyle\alpha\int\lvert g\vert\,d\mu\leq\int f\lvert g\rvert\,d\mu\leq\beta\int\lvert g\rvert\,d\mu$

Now either the integral of $\lvert g\rvert$ is zero or it’s not. If it’s zero, then $g$ is zero a.e., and so is $f\lvert g\rvert$, and our assertion follows for any $\gamma$ we like. On the other hand, if it’s not we can divide through to find

$\displaystyle\alpha\leq\frac{\int f\lvert g\rvert\,d\mu}{\int\lvert g\vert\,d\mu}\leq\beta$

this term in the middle is our $\gamma$.

June 14, 2010 Posted by | Analysis, Measure Theory | 5 Comments