A question concerning measurability of a function 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 :)
 A: 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.
A: 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 :)
