3
$\begingroup$

Let $\Omega\subset \mathbb{R}^n$ be bounded and let $X:=H^1(\Omega)$. Let $a:\Omega\times \mathbb{R} \to \mathbb{R}, (x,z)\mapsto a(x,z)$ be a bounded function such that $a(x,.)$ is continuous on $\mathbb{R}$ for every $x$ and $a(.,z)$ is measurable on $\Omega$ for every $z$.

For $v,\phi\in H^1$ one finds in pde the integral $$\int_\Omega a(x,v(x))\nabla v(x) \cdot \nabla\phi(x) dx$$ All measures should be the Lebesgue measure My question is: Why is $a(x,v(x))$ a measurable function, why is $a(x,z)$ measurable on $\Omega\times \mathbb{R}$ or why is the integral defined?

Usually one can not conclude from measurability of each compontent to measurability on the product space. Somehow i feel I need more measure theory than i know yet :)

$\endgroup$
  • 3
    $\begingroup$ Functions measurable in one coordinate and continuous in the other are Caratheodory functions and they are jointly measurable. $\endgroup$ – Michael Greinecker Mar 15 '13 at 12:11
  • $\begingroup$ thanks, that solves my problem :) Didn't know the name of such functions. $\endgroup$ – Quickbeam2k1 Mar 15 '13 at 12:17
  • $\begingroup$ @Michael: can you post that as an answer? $\endgroup$ – Willie Wong Mar 15 '13 at 12:44
  • $\begingroup$ In a different setting one requires the measurability of $a(x,v)$ where $v \in X:=L^\infty([0,1])$, $a$ still bounded. But $X$ is not separable which was required in your link. Or could one exploit that each such $v$ is in fact in $L^2((0,1))$ and this space is separable? (The integral would then be with $\nabla w$ instead of $\nabla w$ for some $w\in H^1$ (eg for stationary porous medium equation) $\endgroup$ – Quickbeam2k1 Mar 15 '13 at 13:34
1
$\begingroup$

Functions measurable in one coordinate and continuous in the other are Caratheodory functions and they are jointly measurable. This is Lemma 4.51 in Infinite Dimensional Analysis (3rd Ed) by Aliprantis and Border.

$\endgroup$
  • $\begingroup$ Thanks very much. I have one question on that, assume $v(x)\in L^\infty$ (or just a measurable function) then by that theorem it is not clear why the composition $a(x,v(x))$ is a measurable function. The Theorem only states that $a(x,z)$ is measurable. Is such a composition of measurable functions again measurable? (Rudin defines measurability via the preimage of open sets is measurable). So there should be some work left to do? $\endgroup$ – Quickbeam2k1 Mar 15 '13 at 16:06
  • 1
    $\begingroup$ @Quickbeam2k1 Compositions of measurable functions are measurable, but to see that, you need a more general notion of measurability. Rudins notion reduces to the case in which the codomain is endowed with the Borel $\sigma$-algebra. It would probably help if you would familiarize yourself with product $\sigma$-algebras. The chapter on easurability in Aliprantis & Border is probably a good starting point $\endgroup$ – Michael Greinecker Mar 15 '13 at 16:17
  • $\begingroup$ I should be more precise: I know that the composition of measurable functions is again measurable. I searched here also for the compositon on measurable functions and one remark found here link says that the compostion of Lebesgue-measurable functions need not to be measurable again, due to the change of the $\sigma$-algebras (as you also mentioned). But since I want to integrate with respect to Lebesgue I need more Lebesgue-measurability of $a(x,v(x))$. This is not covered by the theorem you mentioned, right? Or am i totallty mistaken? $\endgroup$ – Quickbeam2k1 Mar 15 '13 at 16:39
  • 1
    $\begingroup$ For every Lebesgue measurable function, there is a Borel measurable function that agrees with it almost everywhere. Every Borel-measurable function is also Lebesgue measurable, so you can intgrate them all the same. And for composing, the work much nicr. $\endgroup$ – Michael Greinecker Mar 15 '13 at 17:00
  • 1
    $\begingroup$ The identity function is trivially measurable. The rest should work. $\endgroup$ – Michael Greinecker Jan 4 '14 at 20:17
1
$\begingroup$

Okay, I think I found a more direct approach sufficient to my needs. Whenever, I refer to measurable I mean Lebesgue-measurable on the Lebesgue-measurable set $\Omega \subset \mathbb{R}^n$.

Let $a:\Omega\times\mathbb{R} \to \mathbb{R}$ be a Caratheodory-function, i.e. $a(x,z)$ is measurable in $x$ for every $z$ and continuous in $z$ for every $x$.

Let $v:\Omega \to \mathbb{R}$ be a measurable function. Then $a(x,v(x))$ is mearuable.

$\textbf{Proof:}$ Since $v$ is measurable the exists a simple function $v_k(x)=\sum_{j=0}^k \lambda_j \chi_j(x)\to v(x)$ for every $x$ and with $\lambda_j \in \mathbb{R}$ and $\chi_j$ appropriate characteristic funtions of disjoint measurable sets $E_j$. We assume $\lambda_0=0$ and $E_0=[\cup_{j=1}^n E_j]^C$.

We will show that $a(x,u_k(x))$ is measurable. To this end, we show that $\{x:a(x,u_k(x))>t\}$ is measurable. There holds:

$$\{x:a(x,u_k(x))>t\}=\bigcup_{j=0}^{n}(\{x:a(x,\lambda_j)>t\}\cap E_j)$$ Since $E_j$ is measurable, and $\{x:a(x,\lambda_j)>t\}$ is measurable due to the assumptions we find the assertion.

It remains to pass to the limit $u_k \to u$.

Thanks to the continuity of $a$ in the second argument we find $a(x,u_k(x))\to a(x,u(x))$ for every $x$. On the left hand side we find a measurable function and due to the pointwise limit of measurable functions being measruable again we conclude.

Hope this is error free :)

$\endgroup$

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.