# 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?

• Sep 24, 2014 at 11:35

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.

• Couldnt we give another cover for this...@Yiorgos S. smyrlis Sep 25, 2014 at 10:53

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:

$\lambda_2(l)=\lambda_2(\cup_{k=0}^{\infty}([-k,k]\times 0))\leq\sum_{k=0}^{\infty}\lambda([-k,k]\times 0)=0$

• $[-k,k]\times 0$ is not an interval Jul 31, 2020 at 18:56