refining a partition in $\mathbb R^k$? This should be a simple question, but I searched for a while and couldn't seem to find a good answer.  So I figured I'd try asking it here.
If $X\subset \mathbb R$ is a real interval, $P=\{p_1, p_2, ..., p_m\}$ and $Q=\{q_1, q_2, ..., q_n\}$ are two partitions of $X$ ($p_i, q_j \in \mathbb R$), it's well-known that $R=P\bigcup Q$ is also a partition of $X$, and is, in particular, a refinement of $P$.
How may I extend this language when $X$ is an arbitrary subset of $\mathbb R^k$?  Specifically, suppose $P=\{E_1, E_2, ..., E_m\}$ and $Q=\{F_1, F_2, ..., F_n\}$ are two partitions of $X$, i.e. $\bigcup_{i=1}^m E_i=\bigcup_{j=1}^n F_j=X.$  What I really want is a partition $R$ which refines $P$ by $Q$, i.e. $R=\{G_{ij}\},$ where $G_{ij}=E_i\bigcap F_j,$ with $1\le i\le m, 1\le j\le n$, excluding those $G_{ij}$ which are empty sets.  Is there a compact way to express this?  Thanks a lot!
 A: In this wikipedia article, the following definition is made, which includes nice notation

Further, given two partitions $Q = \lbrace Q_1, \dots, Q_k \rbrace$ and $R = \lbrace R_1, \dots, R_m \rbrace $, we define their refinement as

So instead of ignoring just the empty set, sets of measure $0$ are ignored. The usual measure on $\mathbb{R^n}$ is Lebesgue measure, denoted by $\lambda$. If $P$ and $Q$ consist of Carthesian products of non-degenerate (half-open) intervals, the product of non-degenerate intervals has positive measure, so indeed we are ignoring only empty sets. Also consider the case where 
$X=[0,2]$
$P = \lbrace [0,\frac{1}{2})\cup (\frac{1}{2}, 1) , \lbrace \frac{1}{2} \rbrace \cup [1,2]  \rbrace$
$Q = \lbrace [0,1), [1, 2] \rbrace$
Here, simply intersecting all elements of $P$ and $Q$ would include the singleton $\lbrace \frac{1}{2} \rbrace$ in the refinement. However, this singleton has measure $0$, so instead, we have
$$
P \vee Q = \lbrace [0,\frac{1}{2})\cup (\frac{1}{2}, 1) , [1, 2] \rbrace
$$
Now while this is not really a partition of of $X$, as $\frac{1}{2}$ is not included, it is a partition of a subset $\bar{X}$ of $X$ with full measure, i.e. measure $\lambda(\bar{X}) = 2 = \lambda(X)$. So in a sense $P$, $Q$ and $P \vee Q$ are all equivalent (in the sense of an equivalence class), with $Q$ having the nicest representation.

Edit
The following definition from this wikipedia article may be more relevant to you (especially considering that you applied the tags general-toplogy and elementary-set-theory). The definition involves covers rather than partitions, but of course a partition is a cover. Paraphrasing wikipedia, we get your definition, with the same notation as above.
Let $X$ be a set. For two covers $C$ and $D$ of $X$, we define their minimal common refinement $C \vee D$ as
$$
C \vee D = \lbrace C_i \cap D_j \, | , i \in I, j \in J \rbrace \setminus \lbrace \emptyset \rbrace
$$
The wikipedia article about covers also has a section on refinement. This consistently defines a refinement, in that in this sense $C \vee D$ is indeed a refinement of both $C$ and $D$.
