Lebesgue outer measure in $\mathbb{R}^2$ in terms of a grid of $h$-squares

For a set $D\subseteq\mathbb{R}^2$, the Lebesgue outer measure of $D$ is defined by$$\lambda^\ast(D)=\inf\bigg\{\sum_i\lambda(I_i)\mid D\subseteq\bigcup_iI_i\bigg\},$$where $\{I_i\}$ is a sequence of closed rectangles, and $\lambda(I_i)$ is the area of $I_i$. Now, divide $\mathbb{R}^2$ into a grid of squares with edge length $h$ (called $h$-squares) by the lines $x=0,\pm h,\pm2h,\pm3h,\cdots$ and $y=0,\pm h, \pm2h, \pm3h,\cdots$ Let $D_h$ be the set of $h$-squares intersecting $D$, and$$A(D,h)=\sum_{\sigma\in D_h}h^2.$$If $D$ is compact, then it can be covered by finitely many closed rectangles, the sum of whose areas is arbitrarily close to $\lambda^\ast(D)$, and each rectangle can be approximated by $h$-squares, so\begin{equation}\lambda^\ast(D)=\lim_{h\to0^+}A(D,h).\tag{$\ast$}\label{this}\end{equation}My question is whether \eqref{this} holds even when $D$ is not compact. To make things simpler, let us assume that $D$ is bounded.