Measure of a "tightened" set?

Question

I am trying to see a compact way of proving the following. Let's have a set $F\subseteq \mathbb R^n$ (basic properties can be assumed, e.g. closed, compact, etc), and consider the "tightened" set $$T(\delta) = \left\{x\,|\, B(x,\delta) \subseteq F,\,\, \forall\, x\in F \right\}$$ where $B(x,\delta)$ is a ball centred at $x$ and of radius $\delta$. Then I would like to prove that an inequality of the form: $$\mu(T(\delta)) \geq \mu(F) - c\delta$$ holds for some $c\geq 0$, $\delta$ small enough, where $\mu$ is a measure on $F$ (assume what is needed on $\mu$). This appears like it could be a standard result... Anyone has seen something like this before? It appears obviously true on "simple" sets $F$, and possibly false on pathological ones. Are there simple assumptions, and a simple proof of that result?

Thanks!

I think the comments at Measure of $\varepsilon$- fattening will be of interest. Also of possible interest is To show that the set point distant by 1 of a compact set has Lebesgue measure $0$ and the google search Oleksiv + Pesin + "Hausdorff". — Dave L. Renfro, Sep 08 '20 at 16:28
You should be more precise (beyond "etc") what you regard as basic conditions; otherwise you are inviting easy counterexamples. You should also be precise on what exact statement you want to prove: it's quite unclear how $c$ and $\delta$ are quantified, in what order they are quantified, whether there are any hidden quantified variables, and so on. — Lee Mosher, Sep 08 '20 at 16:33

score 1 · Answer 1 · answered Sep 08 '20 at 16:26

1

There is no $c$ for which such an equality holds.

Consider $F$ a fat Cantor set. This is a compact set which contains no intervals despite having positive measure, so for any $\delta>0$, $T(\delta)$ is empty therefore has measure zero, which rules out such an inequality by considering fat Cantor sets of increasing measure.

Here I am using the standard Lebesgue measure as $\mu$, which is extremely well beheaved as far as measures are concerned, so I doubt there are extra assumptions you can add on $\mu$ to make this work.

answered Sep 08 '20 at 16:26

Alessandro Codenotti

12,285

Thanks! I had that in mind when talking about "pathological" sets, thanks for pointing a specific example. That brings me to the question on "what assumptions are needed for that to work". $F$ connected maybe? – SGros Sep 08 '20 at 16:34
@SGros I'm pretty sure connected is not enough for subsets of $\Bbb R^n$ with $n\geq 2$, convex might work though – Alessandro Codenotti Sep 08 '20 at 16:35
Yeah, I'm afraid connected is not enough. But I take it my question is not a simple one. That's already useful information. – SGros Sep 08 '20 at 16:49

Measure of a "tightened" set?

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