Continuous function over measurable set

Question

Assume we have f : E → R to be a continuous function and $E ⊂ R^n$ be a closed Jordan measurable set.

Show if E is connected and μ(E) > 0, then there is a point c ∈ E s.t. $\frac 1 {μ(E)} \int_E f(x)dx = f(c)$

Is μ(E) the mean of the set? I know this is related to mean value theorem but is the the same proof?

Answer 1

Here, $\mu$ is presumably Lebesgue measure, so $\mu(E)$ is the measure of $E.$

As far as the proof of this statement, it is almost identical. I'm assuming $E$ is bounded, as that's the only class of sets to which I've seen someone apply "Jordan measurable". If so, then $E\subset \mathbb{R}^n$ is closed and bounded, hence compact, and $f(E) \subset \mathbb{R}$ is the continuous image of a compact, connected set, hence it is compact and connected. In particular, this shows that $f(E) = [a,b]$ for some real numbers $a,b.$

From this point, it's effectively the same as the proof of the usual mean value theorem in that it comes from the basic properties of the integral: $$a\mu(E) = \int_E a\,dx \leq \int_Ef(x)\,dx \leq \int_Eb\,dx = b\mu(E)$$ so that $$\frac{1}{\mu(E)}\int_E f(x)\,dx \in [a,b] = f(E)$$ from which we immediately can conclude the result.

Answer 2

$\mu$ is the Lebesgue measure on $\mathbb R^{n}$. To prove this result you only have to note that $\frac 1 {\mu (E)} \int_E f d\mu$ lies between the maximum and the minimum values of $f$ (these could be $\infty$ or $-\infty$) and $f(E)$ is an interval. Hence $f$ attains all values between the maximum and the minmum, in particular the value $\frac 1 {\mu (E)} \int_E f d\mu$.

Answer 3

Since the continuous image of a connected set is connected, $f(E)$ must be an interval. Consider the function $g(y) = f(y)-\frac 1 {μ(E)} \int_E f(x)dx$. In particular note that $ \frac 1 {μ(E)} \int_E g(y) dy = 0$. If $g$ is zero anywhere, we are done. Otherwise, as $f(E)$ is an interval, we have the intermediate value property. So either $g(y)>0$ for all $y$ or $g(y)<0$ for all $y$. WLOG we assume the former. As the Jordan measure $\mu(E)$ is positive, $E$ must have non-empty interior. In particular $g(y) \geq \delta>0$ on some rectangle $R\subset E$, so $\frac 1 {μ(E)} \int_E g(y) dy \geq \delta \frac{\mu(R)}{\mu(E)}>0$, a contradiction. It follows that $g$ is somewhere non-zero on $E$ so we are done.