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?


2 Answers 2


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.

  • 1
    $\begingroup$ Couldnt we give another cover for this...@Yiorgos S. smyrlis $\endgroup$
    – David
    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 $

  • 2
    $\begingroup$ $[-k,k]\times 0$ is not an interval $\endgroup$
    – fdzsfhaS
    Jul 31, 2020 at 18:56

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