Why is $A=\{(x_1,x_2,...,x_n)|\exists_{i\ne j}: x_i=x_j\}$ a null set? Why is $A=\{(x_1,x_2,...,x_n)|\exists_{i\ne j}: x_i=x_j\}$ a null set?
This claim was shown in a solution I ran into, and I don't see how it holds. I try to follow the formal definition of nullity, which is having the possibility to be covered by a collection of open cubes, whose volume is as small as desired. I can't, however, tell how it is achieved here. There seem to be too many options and combinations here. I could use some help here.
 A: HINT:
For $n\ge 2$ denote by   $S_n$ the subset of $\mathbb{R}^n $ consisting of points with the first two coordinates equal. Then with the indentification $\mathbb{R}^n =  \mathbb{ R}^2 \times \mathbb{R}^{n-2} $ we have 
$$S_n = S_2 \times \mathbb{R}^{n-2} $$
Hence it is enough to show that $S_2 \subset \mathbb{R}^2$ is a subset of measure $0$.  In fact, it is enough to find a countable cover of $S_2$ by rectangles with finite total area $A$. Indeed, since $S_2$ is invariant under dilations, for any $\delta > 0$   we can find a cover of total area $\delta^2 \cdot A$.   Now, consider a sequence of positive numbers $d_n$ so that $\sum d_m = \infty$ while $\sum d_m^2 < \infty$ (for instance $d_m = \frac{1}{m}$). Arrange along the diagonal starting from the origin squares with sides $d_1$, $d_2$, $\ldots$, $d_m$,  $\ldots$ going one sequence $NE$, another $SW$. Since $\sum d_m = \infty$ the full line $x_1 = x_2$ gets covered, and the total area of the squares will be $2 \sum d_m^2 < \infty$. 
A: The trick is to find a cover of $A_{j;k} = \{x: x_j = x_k\}$ then we use that $A = \bigcup A_{j;k}$ (so a cover for each of $A_{j;k}$ will together cover $A$).
Now to cover it with open cubes the trick is that we first don't have to use cubes of the same size. It should also be understood that we're allowed to have infinite number of cubes, but when measuring their size we get an infinite sum which should converge.
Now to cover it we just select cubes that get's smaller and smaller the longer from the origin we get in such a way that they sum up finitely (and as small as we desire).
A: As suggested in a comment, the claim follows if we can show that
$$A_{ij} := \{(x_1,\ldots,x_n) \in \mathbb{R}^n; x_i = x_j\}$$
is a null set for all $i \neq j$. Without loss of generality, I'll assume that $i=1$, $j=2$.
Since the countable union of null sets is again a null set, it suffices to show that
$$B_K := \{(x_1,\ldots,x_n); x_1=x_2, \forall j=1,\ldots,n: |x_j| \leq K\}$$
is a null set for any $K \in \mathbb{N}$. To this end, we note that
$$B_K := \bigcap_{k \in \mathbb{N}} \bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-K,K] \times \ldots \times [-K,K]). \tag{1}$$
Indeed: If $x \in B_K$, then $x_1=x_2$ and $|x_j| \leq K$ for all $j=1,\ldots,n$. Since any number $y:=x_1=x_2$ can be approximated by dyadic numbers of the form $\ell 2^{-k}$, we find that $x$ is contained in the right-hand side of $(1)$. On the other hand, if $x \in \mathbb{R}^n$ is contained in the right-hand side of $(1)$, then we can find for any $k \in \mathbb{N}$ some $\ell=\ell(k) \in \mathbb{Z}$ such that $x_1,x_2 \in [\ell 2^{-k},(\ell+1)2^{-k})$. Hence,
$$|x_1-x_2| \leq |x_1-\ell 2^{-k}|+ |x_2-\ell 2^{-k}| \leq 2 \cdot 2^{-k} \xrightarrow[]{k \to \infty} 0,$$
i.e. $x_1 = x_2$. This implies $x \in B_K$.
Now we are ready to show that $B_K$ is a null set. Since the sets $[\ell 2^{-k},(\ell+1)2^{-k})$, $\ell \in \mathbb{Z}$, are disjoint for fixed $k \in \mathbb{N}$, we have
$$\begin{align*} &\quad \lambda \left(\bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right) \\ &=\sum_{\ell=-K 2^k}^{K 2^k}\lambda \left(  ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right) \\ &= (2K)^{n-2} 2^{-2k} \sum_{\ell=-K 2^k}^{K 2^k} 1 \\ &= (2K)^{n-2} 2^{-2k} 2K 2^k. \end{align*}$$
Consequently, we find using the continuity of the Lebesgue measure, i.e.
$$\lambda(B_K) = \lim_{k \to \infty} \lambda \left(\bigcup_{\substack{\ell \in \mathbb{Z} \\ |\ell| \leq K 2^k}} ([\ell 2^{-k},(\ell+1)2^{-k}) \times [\ell 2^{-k},(\ell+1)2^{-k}) \times [-R,R] \times \ldots \times [-R,R]) \right)$$
that
$$\lambda(B_K) = \lim_{k \to \infty} ((2K)^{n-2} 2^{-2k} 2K 2^k) = 0.$$
