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Quick question about inner products and integration:

Assume that we have a function $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ where $\Omega \subset \mathbb{R}^{n}$ such that $a(x,\ldots)$ is continuous on $\mathbb{R} \times \mathbb{R}^{n}$ for almost every $x \in \Omega$ and $a(.,s, \xi)$ is measurable for every $(s, \xi) \in \mathbb{R} \times \mathbb{R}^{n}$ then if we consider: $$ \langle a(x_{0},s,\xi) - a(x_{0},s,\eta); \xi - \eta \rangle < 0$$ for $x_{0} \in \Omega \text{ and } s \in \mathbb{R}, \xi, \eta \in \mathbb{R}^{n}$

Does it then follow that $$\int_{B(x_{0}, \rho)}\langle a(x_{0},s,\xi) - a(x_{0},s,\eta); \xi - \eta \rangle dx = \langle a(x_{0},s,\xi) - a(x_{0},s,\eta); \xi - \eta \rangle|B(x_{0}, \rho)| \text{ ?}$$

Thanks for any help

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Yes. For any measurable set $E$ and any constant $c$ we have $\int_E c\,dx=c|E|$. In your case, $E$ is the ball $B(x_0,\rho)$ and $c$ is $$ \langle a(x_{0},s,\xi) - a(x_{0},s,\eta); \xi - \eta \rangle$$ which is a constant in this context, as it does not involve $x$.

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