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Let $\{x_i\}$ be the sequence of all rational points in $\mathbb R^n$, and let $$B_i=\{x\in\mathbb R^n\mid\|x-x_i\|\leq2^{-i}\},~~E=\bigcup^\infty_{i=0}B_i.$$

We know that the Lebesgue measure of $E$, denoted by $|E|$, is finite and $E$ is dense in $\mathbb R^n$.

Q: Does $\partial E=\mathbb R^n\setminus E$ have the infinite Lebesgue measure?

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    Well, I haven't checked but if as you claim, $|E|$ is finite and $\partial E= \Bbb R^n \setminus E$, then $|\partial E |$ trivially has to be infinite. No? – 2017-01-14
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    @OpenBall Could you show why $|\partial E|$ has no infinite Lebesgue measure – 2017-01-14
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    The symbol $\partial E$ usually denotes the boundary of $E,$ not the complement of $E.$ The boundary of $S$ is defined as $\overline S \cap \overline {\mathbb R^n \backslash S}. $ – 2017-01-14
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    If you replace $B_i$ by the corresponding open ball then the boundary of E equals its complement and its measure is infinity. I don't if this is what DLIN wanted. – 2017-07-26

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In the arithmetic of the extended-non-negative-real-half-line $S=[0,\infty]$ we have $x+\infty=\infty +x=\infty$ for all $x\in S$, and $x+y$ has its usual meaning for $x,y\in [0,\infty).$

Let $M_n$ be Lebesgue measure on $\mathbb R^n.$ For any disjoint Lebesgue-measurable $E,F\subset \mathbb R^n$ we have $M_n(E)+M_n(F)=M_n(E\cup F ).$

So $M_n(E)+M_n(\mathbb R^n$ \ $E)=M_n(\mathbb R^n)=\infty.$ So if $M_n(E)<\infty$ then $M_n(\mathbb R^n$ \ $E)$ cannot be finite.