The Campanato space is \begin{align} \mathcal L^{p,\lambda}(\Omega):=\{u\in L^p(\Omega)\colon[u]_{p,\lambda}<\infty\} \end{align} where \begin{align} [u]^p_{p,\lambda}=&\sup_{x_0\in\Omega,r>0}\frac{1}{r^\lambda}\int_{B_r(x_0)\cap\Omega}|u-u_{x_0,r}|^pdx\\ u_{x_0,r}=&\frac{1}{|B_r(x_0)\cap\Omega|}\int_{B_r(x_0)\cap\Omega}udx \end{align}
We can define a new semi-norm by \begin{align} [u]^p:=\sup_{B_r(x_0)\subset\Omega}\frac{1}{r^\lambda}\int_{B_r(x_0)\cap\Omega}|u-u_{x_0,r}|^pdx \end{align} that is we only consider the ball contained in $\Omega$, will this be equivalent to $[u]_{p,\lambda}$?
No, considering only the balls contained in $\Omega$ gives a weaker seminorm. For example, let's consider two-dimensional domains using complex notation: $$\Omega=\{z\in\mathbb{C}: 1<z<2, z\notin (-2, -1)\}$$ The function $u(z)=\arg z$ is Lipschitz on any disk contained in $\Omega$, so whichever $p$ is used (say $p=2$) we get nice decay of the integrals $\int_{B_r(x_0)\cap\Omega}|u-u_{x_0,r}|^2dx$, something like $r^4$ (because the squared difference $|u-u_{x_0,r}|^2$ is $O(r^2)$). Thus $\lambda=4$ can be used here.
But if the ball is centered on the cut $(-2, -1)$, the situation is quite different. Now the function jumps by $2\pi$ on two sides of the cut, hence $|u-u_{x_0,r}|$ is about $\pi$, and the integral $\int_{B_r(x_0)\cap\Omega}|u-u_{x_0,r}|^2dx$ is of order $r^2$. Thus, allowing balls that are not contained in $\Omega$ results in $\lambda \le 2$.