Formally proving that the image of $f(E)$ is of measure zero when $E$ is a null set and f is $C^1$

Question

So $f:U \rightarrow R^n$ is a continuously differential function, and $U \subset R^n$ is an open set. When $E$ a set of measure zero, I need to prove that $f(E)$ is also of measure zero.

I know that this question was asked here before, but I still can't formalize my proof because I am missing something.

I know that since $E$ is of measure zero then $E \subset \bigcup_{i}Q_i$ where $Q_i$ are intervals and $\sum_{i}|Q_i| < \epsilon$ for $\epsilon > 0$.

I also know that since $U$ is an open set there exist intervals $R_i$ that satisfy $U = \bigcup_{j}R_j$.

Also I know that if I look at $E \bigcap ([-k,k] $ x ... x $[-k,k])$ then $f$ is lipshitz on this domain, let's call it $f_k$, and $f(E) = \bigcup_{k}E(f_k)$.

But still I am missing here something that will prove that $f(E)$ is covered by intervals with measure zero each.

Answer 1

This is essentially the proof that Rudin gives in Real & Complex Analysis (Lemma 7.25).

For $k,\ell \in \mathbb N$, suppose that $$F_{k,\ell} = \left\{x \in E \, : \, \frac{\lvert f(x) - f(y)\rvert}{\lvert x - y \rvert} \le k, \,\,\, \forall y \in B(x,1/\ell)\cup E\right\}.$$ Because $C^1$-functions are locally Lipschitz, we have $$E = \bigcup_{k,\ell \in \mathbb N} F_{k,\ell}.$$ Since $$f(E) = \bigcup_{k,\ell \in \mathbb N} f(F_{k,\ell}),$$ it suffices to show that $f(F_{k,\ell})$ has measure zero for all $k,\ell \in \mathbb N$. Since $F_{k,\ell} \subset E$, we see that $\lvert F_{k,\ell} \rvert = 0$ and thus for any $\epsilon > 0$, we can find a countable collection of balls $B_i = B(x_i,r_i)$ where $x_i \in F_{k,\ell}$, $r_i < 1/\ell$ and $$\sum^\infty_{i=1} \lvert B_i \rvert \le \frac{\epsilon}{k^n}.$$ Now if $x \in F_{k,\ell} \cap B_i$, then $\lvert x - x_i \rvert < r_i$, which shows that $$\lvert f(x) - f(x_i) \rvert \le k \lvert x - x_i\rvert < kr_i.$$ Thus $f(F_{k,\ell} \cap B_i) \subset B(f(x_i),kr_i).$ This shows that $$\lvert f(F_{k,\ell}) \rvert \le \sum_{i=1}^\infty \lvert f(F_{k,\ell} \cap B_i) \rvert \le \sum^\infty_{i=1} \lvert B(f(x_i),kr_i)\rvert = k^n\sum^\infty_{i=1}\lvert B_i \rvert \le \epsilon.$$ Since this holds for arbitrary $\epsilon > 0$, this shows that $\lvert f(F_{k,\ell}) \rvert = 0$ and completes the proof.

Answer 2

We will prove that for every natural number $N$, $f(E\cap \overline{B_N(0)})$, where $\overline{B_N(0)}$ is the closed ball of center at the origin and radius $N$, is measurable and it has measure zero. This implies that $f(E)$ has measure zero because $f(E)=\cup_{N\in\mathbb{N}}f(E\cap \overline{B_N(0)})$ and we can apply the $\sigma$-subadditivity property. In other words, we reduce the problem to the case of a measure zero and bounded set.

Let $N$ be a natural number and consider the closed ball $\overline{B_N(0)}$, which is compact. So from the Mean Value Theorem (for several variables) it follows that $f$ is Lipschitz on $\overline{B_N(0)}$ with Lipschitz constant $C:=\max \{||Df(x)z||:x\in \overline{B_N(0)}, z\in \mathbb{R}^n \text{ with } ||z||=1\}$.

Thus for every subset $A\subseteq \overline{B_N(0)}$, we have that $|f(A)|_o\leq C|A|_o$, where $|\cdot|_o$ denotes the Lebesgue's outer measure. (Try to check that and let me know if you have some difficulty.) In particular, we have that $|f(E\cap \overline{B_N(0)})|_o\leq C|E\cap \overline{B_N(0)})|_o\leq C|E|_o=0$. So $f(E\cap \overline{B_N(0)})$ is measurable and it has measure zero.

By the argument of the first paragraph, the proposition follows.