Let $N>1$ be a natural number. Let $C:\mathbb R^N\times\mathbb R^N\rightarrow\mathbb R$ be a $L^1(\mathbb R^N\times\mathbb R^N)$ function such that $C(x,y)=C(y,x)$ (this is a condition that probably will not be necessary, but come as bonus). Denote $x\cdot y$ the usual inner product in $\mathbb R^N$. For each $R>0$ denote $\mathcal X_R:\mathbb R^N\rightarrow\mathbb R$ the characteristic function of the open ball $B_R(0)=\{x\in\mathbb R^N;\ \!|x|<R\}$, where $|x|=(x\cdot x)^{1/2}$ is the usual euclidean norm in $\mathbb R^N$. That means: $$ \mathcal X_R(x)=\left\{\begin{array}{cc} 1\ \ \mbox{if}\ \ x\in B_R(0);\\ 0\ \ \mbox{if}\ \ x\notin B_R(0). \end{array}\right. $$ Can we prove that $$ \lim_{R\rightarrow+\infty}\iint_{\mathbb R^N\times\mathbb R^N}C(x,y)\frac{(x\mathcal X_R(x)-y\mathcal X_R(y))\cdot(x-y)}{|x-y|^2}dxdy=\iint_{\mathbb R^N\times\mathbb R^N}C(x,y)dxdy\ \ ? $$
The pointwise convergence of the integrand is evident but I cannot get a suitable pointwise boundedness to use Lebesgue's Dominated Convergence Theorem. Notice that whenever $x$ and $y$ are close enough but one of them is in the ball and the other is not, we have problem. Nevertheless, it seems quite reasonable that this limit should hold. If it makes it easier, we can change general $R$ by natural numbers $n$, to work with the limit $n\rightarrow\infty$.
