Can $\mathbb{R}^{n}$ be expressed as the union of disjoint positive measure sets with cardinal $\aleph_1$ As the title suggested, can we find positive measure sets$\{V_j\}_{j\in\alpha}$, such that $$V_j\cap V_k=\emptyset, \quad \cup_{j\in\alpha}{V_j=\mathbb{R}^{n}},$$
and the cardinal number of this set is $\aleph_1$.
 A: Edit: This answer is now obsolete following an edit to the question; but hardmath has posted a very nice solution to the intended question.

By $\aleph$ do you mean $\aleph_0$? If so then yes: for instance, define
$$V_k = [k,k+1) \times \mathbb{R}^{n-1} \subseteq \mathbb{R}^n$$
Then $\mathbb{R}^n = \displaystyle \bigcup_{k \in \mathbb{Z}} V_k$ and $V_k \cap V_{\ell} = \varnothing$ for $k \ne \ell$. (Imagine chopping $\mathbb{R}^n$ into layers.)
In fact you can express $\mathbb{R}^n$ as a disjoint union of countably many sets of positive finite measure using intuition along similar lines: take unit $n$-cubes instead of layers.
A: If $\mathbb{R}^n$ is the disjoint union of subsets of positive measure, there are at most $\aleph_0$ of these subsets.
Suppose that there are an infinite number, as otherwise the proposition is evident.  Partitioning any subsets of infinite measure into ones of finite measure can then be done without increasing the (infinite) cardinality of these subsets, as infinite measured portions are subdivided into countably many pieces, e.g. intersections with half-open hypercubes $\Pi[k_i,k_i+1)$.
Now count for each $1/r$, $r = 1,2,\ldots$, how many such subsets have measure at least $1/r$.  There are at most countably many.  Summing the count (including repetitions) as $r \rightarrow \infty$ shows that there are at most countably many of positive measure.
I think I remember an exercise like this in Rudin.
