Possible areas for convex regions partitioning a plane and containing each a vertex of a square lattice.

Question

If the plane is partitioned into convex regions each of area $A$ and each containing a single vertex of a unit square lattice, is $A\in (0,\frac{1}{2})$ possible?

If each each vertex is in the interior of its region is $A \neq 1$ possible?

More generally if $\rm{ I\!R}^n$ ($n\ge 1$) is partitioned into convex regions, each of $n$-volume $A$ and each containing a single vertex of a unit $n$-cube lattice and letting $L_n$ be the infimum of possible values of $A$:

For $m>n,\ L_m\leq \frac{n!}{m!}L_n$, and for $n=1,\ A=L_1=1$. What are the remaining values of $L_n$ and is each value of $A$ in $[L_n,\infty)$ possible? Failing which in $(L_n,\infty)$? (Note that in view of the response to Del's comment below, for the purposes of my question $\infty$ can be replaced by $\frac{1}{n!}$ in the given ranges.)

Again, if each each vertex is in the interior of its region is $A \neq 1$ possible?

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Small print if any clarification is needed.

A partition of the plane is a set of pairwise disjoint subsets whose union is the whole plane.

A region $R$ is convex if for any points $X,Y\in R$ and any point $Z$ of $XY$, $Z\in R$.

My use of the word "region" here is as in Apostol, Mathematical Analysis.

A set in $R^n$ is called a region if it is the union of an open connected set with some, none, or all of its boundary points.

The term "area" (resp. "$n-$volume") can be taken as Lebesgue measure (https://en.wikipedia.org/wiki/Lebesgue_measure). The values $0$ and $\infty$ are acceptable in the measure, but not possible in a partition as described.

The vertices of a unit square (resp. unit $n-$cube) lattice are the points $\sum_{i=1}^nn_i{\mathbf b}_i$ for some orthonormal basis $\{{\mathbf b}_i\}$ and choice of origin, with $n_i\in \mathbb{Z}$.

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Response to comment by Del below.

The construction is recursive. The recursion is on the number of dimensions $n$ of $\rm{ I\!R}^n$.

A schematic is shown below. The figure is in $\rm{ I\!R}^{n+1}$ and the rectangles in the $k$-flat $x_{n+1}=i$ are meant to represent individual regions in a partition satisfying the conditions in $\rm{ I\!R}^n$. (The rectangles are purely schematic and not intended to indicate anything about the shapes of the regions.) A point $O$ is chosen from the region $\{-\infty

The $(n+1)$-volumes of the resulting regions decrease from $A$ to $A/(n+1)$ as $O$ moves from $+\infty$ to the the $n$-flat $x_{n+1}=i+1$ and increase without bound as $O$ approaches the $n$-flat $x_{n+1}=i$ from below. If $O$ is between the two planes the construction doesn't result in regions because the interiors of the figures formed become disconnected.

This shows both that $L_n\leq\frac{1}{n!}$, and that for $m>n, L_m\leq\frac{n!}{m!}L_n$.

Notice that the induction step will always result in a partition in $\rm{ I\!R}^{n+1}$ with the lattice vertices on the boundary of their corresponding regions irrespective of whether the same was true of the partition in $\rm{ I\!R}^n$ from which it was derived. A vertex cannot be moved into the interior by a non integral translation in the $x_{n+1}$ direction without breaking the 1-1 correspondence between regions and vertices (except when a parallel projection is used). Hence parts two and four of the question.

The two figures below show firstly a possible result of applying the construction to produce a partition in $\rm{ I\!R}^2$ from $\rm{ I\!R}^1$ and then the same result after a unimodular transformation. The intervals $[a_i,a_{i+1})$ in the first diagram represent a conforming partition (necessarily $A=1$) of $\rm{ I\!R}^1$ and the projection points are on $y=i+2$ resulting in half unit intervals between the $b_i$. The area of the regions (in each figure) is thus $\frac{3}{4}$.

The next induction step described above could use either figure, and there are many ways of cutting and pasting bits from the same or transformed figures to give new partitions, but using the above procedure these do not result in any new values for $A$.

It is trivial to prove that $A=0$ is impossible. Slightly less trivial to prove, but true, that $A=\infty$ is impossible. So the first and third parts of the question are asking about values in $(0,\frac{1}{n!})$.