measurability with zero measure

Let $f : [0,1] \rightarrow \Bbb R$ is arbitrary function , and $E \subset \{ x \in [0,1] | f'(x)$ exists$\}$.

How to prove this statement:

If $E$ is measurable with zero measure then $f(E)$ is measurable with zero measure.

If $E$ is measurable with zero measure then $m(E)=0$ that $m :$ {measurable} $\rightarrow \Bbb R$, $m^* : P(\Bbb R) \rightarrow \Bbb R$,$A:= \{ x \in [0,1] | f'(x)$ is be exist$\}$. so $m(E)=m^*(E) <= m^*(A)$

Solution 1:

WLOG, $|f'(x) |< 1$ for all $x \in E$. Fix $\varepsilon > 0$.

Let $U \supseteq E$ be an open set of measure less than $\varepsilon$. Note that for each $x \in E$, there are arbitrarily small intervals $I_x \subseteq U$ centered at $x$ that satisfy:

($\star$): For all $y \in I_x$, $|f(y) - f(x)| < |y - x| < |I_x|/2$. In particular, $f[I_x]$ is contained in an interval centered at $f(x)$ of length $\leq |I_x|$.

Now consider the collection $\mathcal{V}$ of all intervals $J$ centered at $f(x)$ for some $x \in E$, such that for some $I_x \subseteq U$ satisfying ($\star$), we have $f^{-1}[J] \supseteq I_x$ and $|J| \leq |I_x|$. $\mathcal{V}$ is a Vitali covering of $f[E]$ - This means that for each $y \in f[E]$, for each $\delta > 0$ there is an interval $J$ in $\mathcal{V}$ such that $|J| < \delta$ and $y \in J$. By Vitali covering theorem, there is a disjoint collection $\{J_n : n \geq 1\} \subseteq \mathcal{V}$ which covers all but a measure zero part of $f[E]$. Since the $I_x$'s corresponding to these $J_n$'s are pairwise disjoint, the sum of the lengths of $I_x$'s (which is no less than the sum of the lengths of $J_n$'s) is at most measure of $U$ which is less than $\varepsilon$. Letting $\varepsilon$ go to zero, it follows that $f[E]$ has measure zero.

Solution 2:

hot_queen's answer looks fine to me. This is not an independent answer, it's a minor variation and a comment on hot_queen's answer; I'm posting it as an answer for better readability (and who knows, someone might be fooled into upvoting it).

I'm reorganizing the covering lemma part of the argument. As observed earlier, it suffices to discuss the situation where $|f'|\le 1$ on $E$. For each $x\in E$, we can then find an interval $I_x=(x-h_x, x+ h_x)$ so that $|f(I_x)|\le 2 |I_x|$. We can also demand that $I_x\subset U$ for some open set $U\supset E$ with $|U|<\epsilon$.

The intervals $I_x$ ($x\in E$) cover $E$, and we can find a subcollection $I_j$ (necessarily countable) that still covers $E$ and has overlap at most $2$: $\sum \chi_{I_j}\le 2$. This is a version of Besicovich's covering lemma; in one dimension, this is quite easy and immediately plausible. Now $$ |f(E)|\le \sum |f(I_j)| \le 2 \sum |I_j| \le 4 \left| \bigcup I_j \right| < 4\epsilon , $$ as desired.

Solution 3:

Hint. Let $E_n=\{x\in E:|f'(x)|\le n\}$. Prove that each $f(E_n)$ has measure $0$.

Comment. The above hint is perhaps not very helpful. I realized I do not know how to show that $m(f(E))=0$ even is I assume that, say, $|f'(x)|\le\frac13$ for all $x\in E$. On the other hand if one could solve the problem under this additional assumption, then one would also be able to prove in general that $m(f(E_n))=0$ for each $n$, where $E_n$ are as defined in the hint. Indeed we could define, say, $g(x)=\frac1{3n}f(x)$, and then $|g'(x)|\le\frac13$ for each $x\in E_n$. If we knew that $m(g(E_n))=0$ then clearly we would also have $m(f(E_n))\le3n\cdot m(g(E_n))=0$. The question has not yet been answered.

Edit. I do not follow my own hint. I know that in addition to, say, $|f'(x)|\le\frac13$ for all $x\in E$, after throwing out at most countably many points from $E$, we may also assume that for each $x\in E$ and each $t>0$ both $(x-t,x)\cap E$ and $(x,x+t)\cap E$ are uncountable. This might be relevant to help cover $E$ with certain open intervals ... but it is getting a bit vague, and I have not done it, and there might be something easier or better.