1
$\begingroup$

According to first mean value theorem for integration, if $G \ : \ [a,b] \to \mathbb{R}$ is a continuous function, there exists $x \in (a,b)$ such that $$\int_a^b G(t) dt = G(x)(b-a)$$

Assume $G$ is a continuous function defined on $[a,b]$. For $0 < h < \frac{b-a}{2}$ $$\overline{G} \ : \ y \mapsto \int_{y-h}^{y+h} G(t) dt$$ is defined for $y \in [a+h,b-h]$. Applying the first mean value theorem for integration, for all $y \in [a+h,b-h]$, there exists $c_y \in (y-h,y+h)$ with $$\overline{G}(y)=\int_{y-h}^{y+h} G(t) dt = 2 h G(c_y)$$

Taking for $G$ a constant function, $c_y$ can by any point in $(y-h,y+h)$. Hence we can pick up it in a way for which $y \mapsto c_y$ is not a Lebesgue measureable function.

Question: can one find a continuous function $G$ for which $c_y$ is defined for all $y \in (a+h,b-h)$ as the lower bound of the $z$ such that $\int_{y-h}^{y+h} G(t) dt = 2 h G(z)$ and such that $y \mapsto c_y$ is not Lebesgue measureable?

$\endgroup$
  • $\begingroup$ Non-Lebesgue measurable, no. Discontinuous, yes. $\endgroup$ – Gerald Edgar Jun 7 '15 at 22:31
3
$\begingroup$

This is not quite an answer, but I'm going to post it anyway because I think it may lead to a real answer.

If we modify your question so that instead of $c_y$ we consider $$d_y = d(y) = \inf\left\{z \in [y-h, y+h] : \int_{y-h}^{y+h} G(t)\,dt = 2 h G(z) \right\}$$ (note the closed interval instead of open), then we can show that the function $d$ is lower semicontinuous, and in particular is Lebesgue measurable. Fix a sequence $y_n \to y_0$ and set $d_0 = \liminf_{n \to \infty} d(y_n)$. We have to show $d(y_0) \le d_0$. Passing to a subsequence, we may assume $d_0 = \lim_{n \to \infty} d(y_n)$. Note that $y_n -h \le d(y_n) \le y_n + h$ by definition, so $y_0 - h \le d_0 \le y_0 + h$; in particular $d_0$ is finite.

Now for each $n$, by definition of $d$ there exists $z_n$ such that $d(y_n) \le z_n \le d(y_n) + \frac{1}{n}$ and $$\int_{y_n-h}^{y_n+h} G(t)\,dt = 2 h G(z_n). \tag{*}$$ By the squeezing lemma, $z_n \to d_0$ and so by continuity of $G$ we have $2hG(z_n) \to 2hG(d_0)$. On the other hand, by the dominated convergence theorem we have $$\int_{y_n-h}^{y_n+h} G(t)\,dt \to \int_{y_0-h}^{y_0 + h} G(t)\,dt.$$ (To see this, set $F_n = 1_{[y_n - h, y_n + h]}$ and note that $F_n(t) \to F_0(t)$ for all $t \ne y_0 \pm h$; in particular $F_n \to F_0$ almost everywhere. So $F_n G \to F_0 G$ almost everywhere, and is dominated by $G$ on some bounded interval large enough to contain all $[y_n - h, y_n + h]$.)

So when we pass to the limit in (*), we obtain $$\int_{y_0 - h}^{y_0 + h} G(t)\,dt = 2 h G(d_0).$$ Therefore, $d(y_0) \le d_0$.

If we work with $c(y)$ instead, most of the argument above still works. But the very last line could fail because we might have $c_0 = y_0 - h$ in which case we cannot conclude $c(y_0) \le c_0$. I think there should be a way to fix this, but it's too late at night here to keep thinking about it...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.