A Lipschitz transform maps measurable set to measurable

Question

Prove that a Lipschitz transform $T: \mathbb{R}^n \to \mathbb{R}^n$ maps measurable set to measurable.

Assume the only thing that we know about Lipschitz transform is that we can find $M>0$ such that for any $x,y \in \mathbb{R}^d$,

$|T(x)-T(y)|\leq M |x-y| \tag{1}$

To use any property of Lipschitz transforms rather than above, we have to prove it.

We can prove it as follows:

1. $T(\cdot)$ maps every Null set to a Null set (I proved).
2. $T(\cdot)$ maps every $F_{\sigma}$ set to a $F_{\sigma}$ set.
3. Every measurable set can be written as union of a $F_{\sigma}$ and a Null set. (I proved)

I proved step 2 when $m(F_{\sigma})<\infty$. For a general $m(F_{\sigma})$, we have to first show that the $F_{\sigma}$ can be written as countable union of compact sets, I don't know how to do. Also, the map of each compact set is compact, which I don't know how to prove using just using property (1). Any idea?

(I found a similar question-solution that doesn't address my needs here $f$ maps measurable sets to measurable sets)

Answer 1

Let $F \in F_{\sigma}$, then $F = \bigcup_{n = 1}^{\infty}C_n$, where $C_n$ is a closed set. Let $B_n = \overline{B(0,n)}$, then we have that

\begin{align} F = &\ (C_1 \cap B_1) \cup \big((C_1 \cup C_2) \cap B_2\big) \cup \big((C_1 \cup C_2 \cup C_3) \cap B_3\big) \cup \dots \\ = &\ \bigcup_{n = 1}^{\infty}\Big(\big(\cup_{i = 1}^n C_i\big) \cap B_n\Big), \end{align} which expresses $F$ as the countable union of compact sets.

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To prove that a Lipschitz function maps compact sets into compact sets it is enough to notice that $(1)$ implies continuity. The image of compact set under a continuous function is compact. Indeed let $K \subset \mathbb{R}^n$ be compact and let $\mathcal{U}$ be an open cover of $T(K)$. By continuity of $T$ we have that $\{T^{-1}(U) : U \in \mathcal{U}\}$ is an open cover of $K$, hence it admits a finite subcover $\{T^{-1}(U_1), \dots, T^{-1}(U_N) \}$. This shows that $$T(K) \subset T\Big(\bigcup_{i = 1}^NT^{-1}(U_i)\Big) \subset \bigcup_{i = 1}^NU_i,$$ hence $T(K)$ is compact.

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The following proof uses the Lipschitz condition $(1)$ more explicitly.

A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. Let $K$ be compact and consider $T(K)$. Let $R$ be such that $K \subset B_R$ (we can find $R$ since $K$ is compact and hence bounded). Then we have that $$\sup_{(x,y)}|T(x) - T(y)| \le \sup_{(x,y)}M|x - y| \le 2MR.$$ This proves that any two point in $T(K)$ are no more than $2MR$ apart, hence showing that $T(K)$ is bounded. To prove that $T(K)$ is closed let $\{y_n\}$ be a sequence in $T(K)$ and assume that $y_n \to y.$ We need to prove that $y \in T(K)$. For every $n$, let $x_n$ be such that $y_n = T(x_n)$. Now $\{x_n\}$ is a sequence in the compact set $K$, hence there are a subsequence $\{x_{n_k}\}$ and $x \in K$ such that $x_{n_k} \to x$. Then we have:

$$ |y - T(x)| = \lim_{n_k}|y_{n_k} - T(x_{n_k})| = \lim_{n_k}|T(x_{n_k}) - T(x_{n_k})| = 0.$$ This proves that $y = T(x)$ and hence $T(K)$ is also closed.