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Let $X \subseteq \mathbb{R}^n$ be Lebesgue measurable*, and suppose $f : X \to \mathbb{R}^n$ has the property that $d(f(x), f(y)) \le d(x, y)$ for all $x, y \in X$. Suppose furthermore that $f(X)$ is measurable. Must $\mu(f(X)) \le \mu(X)$?

Also, if we relax the conditions that X and $f(X)$ are measurable and consider only the outer measures, must $\mu*f(X) \le \mu*(X)$?

If you know the answer to either of these, that would be helpful.

*This is the " standard" Lebesgue measure, where $\mu(X)$ is the infimum of the sum of the volumes of boxes containing X, the volume of a box being the product of its dimensions.

Perry Ainsworth
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    Depends on the measure $\mu$. Consider a translation and $\mu$ the Dirac measure. On the other hand, this is true if $\mu$ is any Hausdorff measure, almost by definition of such measures. – Jose27 Dec 16 '21 at 03:44
  • @Jose27 I am referring to a specific measure in my question; I believe it is called the Lebesgue measure. It is where you take the infimum of the sum of the volume of boxes containing X, the volume of a box being the product of its dimensions. – Perry Ainsworth Dec 16 '21 at 03:56
  • Yes: https://math.stackexchange.com/questions/158029/measure-of-image-of-lipschitz-function-is-bounded – Jose Avilez Dec 16 '21 at 04:00
  • @JoseAvilez Does that really answer the question though? In the question given, the domain of f is all of $\mathbb{R}^n$, whereas here it is only a subset. If every "shrinking" function can be extended to all of $\mathbb{R}^n$, that would solve this. – Perry Ainsworth Dec 18 '21 at 00:21
  • Lipschitz functions can be extended, per the Kirszbraun theorem. – Michael Jesurum Dec 21 '21 at 19:51

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