Lipschitz image $mathbf{R} to mathbf{R}^2$ has measure zero?
$begingroup$
I know that of a proof of the following fact using Hausdorff measure, but is there a more elementary way to do it?
Suppose $f: mathbf{R} to mathbf{R}^2$ is Lipschitz. Then $mathcal{L}^2(f(mathbf{R})) = 0$.
Above, $mathcal{L}^2$ is Lebesgue outer measure on $mathbf{R}^2$.
measure-theory geometric-measure-theory
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
$endgroup$
add a comment |
$begingroup$
I know that of a proof of the following fact using Hausdorff measure, but is there a more elementary way to do it?
Suppose $f: mathbf{R} to mathbf{R}^2$ is Lipschitz. Then $mathcal{L}^2(f(mathbf{R})) = 0$.
Above, $mathcal{L}^2$ is Lebesgue outer measure on $mathbf{R}^2$.
measure-theory geometric-measure-theory
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
$endgroup$
add a comment |
$begingroup$
I know that of a proof of the following fact using Hausdorff measure, but is there a more elementary way to do it?
Suppose $f: mathbf{R} to mathbf{R}^2$ is Lipschitz. Then $mathcal{L}^2(f(mathbf{R})) = 0$.
Above, $mathcal{L}^2$ is Lebesgue outer measure on $mathbf{R}^2$.
measure-theory geometric-measure-theory
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
$endgroup$
I know that of a proof of the following fact using Hausdorff measure, but is there a more elementary way to do it?
Suppose $f: mathbf{R} to mathbf{R}^2$ is Lipschitz. Then $mathcal{L}^2(f(mathbf{R})) = 0$.
Above, $mathcal{L}^2$ is Lebesgue outer measure on $mathbf{R}^2$.
measure-theory geometric-measure-theory
measure-theory geometric-measure-theory
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
asked Jan 8 at 14:13
Drew BradyDrew Brady
731315
731315
add a comment |
add a comment |
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