Area-minimising hypersurface with unbounded area growth Let $T$ be an $n$-dimensional area-minimising hypersurface in $\mathbf{R}^{n+1}$. If $T$ has bounded area growth in the sense that there is a constant $C > 0$ so that $\mathcal{H}^n(T \cap B_R) \leq C R^n$ for all $R > 0$, then there are rigidity theorems for $T$. For example, when $n \leq 6$ then the work of Simons [1] implies that $T$ must be an $n$-dimensional plane. (In larger dimensions there are singular area-minimising hypercones.)
Question. Is there an example of an area-minimising hypersurface with unbounded growth? Could such an example exist in low dimensions, when $n \leq 6$? What about $n = 2$?
[1] James Simons. Minimal varieties in Riemannian manifolds. Annals of Mathematics, Second Series, Vol. 88, No. 1 (1968), pp. 62-105.
 A: This is a straightforward comparison argument.  Let $\omega_n$ be the volume of $\partial B_1\subset \mathbb{R}^{n+1}$.
For generic $R$, one has $\partial B_R \cap T=\tau$ a smooth submanifold.  By Alexander duality, there is a subset, $\Omega$, of $ \partial B_R\setminus \tau$ so $\partial \Omega=\tau$.  Clearly, $\mathcal{H}^n(\Omega)\leq \omega_n R^n$. In fact, up to replacing $\Omega$ by it's complement one has $\mathcal{H}^n(\Omega)\leq \frac{1}{2}\omega_n R^n$.
By the area minimization property,
$$\mathcal{H}^n(\Sigma\cap B_R)\leq \mathcal{H}^n(\Omega)\leq \omega_n R^n.$$
The monotonicity formula ensures the bound holds for all $R$.
In fact, this argument should work (using slicing) for an area minimizing integral $\mathbb{Z}_2$ current.
A: When $n = 2$ the sort of examples you require does not exist. This is due to Fischer-Colbrie and Schoen, "The structure of complete stable minimal surfaces in 3-manifolds of non-negative scalar curvature".
https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160330206
Essentially: requiring just that the minimal surface is area minimizing for compact perturbations up to second order (which is weaker than the strict area-minimizing condition you asked for), they prove (among other things) that the only "stable" minimal surface in $\mathbb{R}^3$ is the plane.
The reason is basically that the second variation of the area gives a Laplace equation with a potential, which in $n = 2$ can be related to the scalar curvature of the minimal surface. And non-trivial minimal surfaces in $\mathbb{R}^3$ all have negative scalar curvature.
