Lebesgue measure of any line in $\mathbb{R^2}$. What is the Lebesgue measure of a line in $\mathbb R^2$? I am guessing that this zero. But I couldn't prove it rigorously. Please help.
From this can I conclude that any proper subspace of $\mathbb R^n$ has measure zero?
 A: If the line is
$$
\ell=\{(x,0): x\in\mathbb R\},
$$
then for every $\varepsilon>0$, we have
$$
\ell\subset \bigcup_{k\in\mathbb Z}I_k^\varepsilon,
$$
where
$$
I_k^\varepsilon=[k,k+1]\times[2^{-|k|-2}\varepsilon,-2^{-|k|-2}\varepsilon].
$$
But
$$
m_2(\ell)\le\sum_{k\in\mathbb Z} m_2(I_k)=\sum_{k\in\mathbb Z} 2^{-|k|-1}\varepsilon=\varepsilon.
$$
Thus $m_2(\ell)<\varepsilon$, for every $\varepsilon>0$, and hence $m_2(\ell)=0$.
Any other straight line in $\mathbb R^2$ is obtained by a rigid motion of $\ell$, and rigid motion does not change the Lebesgue measure of a set.
Indeed, every proper linear subspace (or more generally, every proper hyperplane) 
of $\mathbb R^n$ has zero $n-$dimensional Lebesgue measure.
A: Without loss of generality  we can suppose that over line $l$ has the form $l=\mathbb R \times 0$. 
This follows from the translation- and rotation invariance of the lebesgue-measure.
(We can move our line such that it contains zero. Then we can rotate it)
We write $l=\cup_{k=0}^{\infty}([-k,k]\times 0)$
Now we obtain:
$\lambda_2(l)=\lambda_2(\cup_{k=0}^{\infty}([-k,k]\times 0))\leq\sum_{k=0}^{\infty}\lambda([-k,k]\times 0)=0 $
