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.
Best Answer
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.