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I am working on the following problem:

Suppose $C \subset \mathbb{R}^d$ is a compact and non-empty set. Let $C_0 = C$ and let $C_t = \{x \in \mathbb{R}^d : d(x,C) \leq t \}$ for all $t >0$. Also, let $g(t) = m(C_t)$, where $m$ is the Lebesgue measure.

(i) Prove that g is right continuous.

(ii) Prove that for $t > 0$, the set $\{x \in \mathbb{R}^d : d(x,C) = t \}$ has Lebesgue measure $0$.

(iii) Using part (ii), show that $g$ is also left continuous.

For part (i), I am thinking to use the fact that the limit of $m(C_t)$ as $T$ goes to 0 is $m(C)$ (I believe that this is true, for example see a very similar claim in Stein and Shakarchi, Measure Theory, chapter 1 exercise 5). Also, for part (ii) I believe that a sketch of the proof is the following: consider a point $x_0$ such that $d(x_0,C)=t$ and then show that the density of ${x: d(x_0,C)<t}$ at a is $\geq 1/2$. We then need to apply the Lebesgue Differentiation Theorem to ${ x: d(x,C)=t }$. I am not sure how to complete this argument. Thank you.

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up vote 2 down vote accepted

Let $x_n$ be a sequence converging monotonically to $t$ from the right. Then, $$C_t = \bigcap_n C_{x_n}.$$ By continuity of measure we then have $$g(t) = m(C_t) = \lim_{n \rightarrow \infty} m(C_{x_n}) = \lim_n g(x_n).$$

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