Prove that $H=\{1/x|x \in E\}$ has measure zero if $E \subset (0,1)$ does.

Question

I guess it's true for functions that are Lipshitz or uniformly continuous since we can limit the length of the intervals after the transformation.
However, I don't know if it's true or not, and since $1/x$ is not one of those, I don't know how to solve this problem.

Answer 1

Here is an interesting result that helps with your problem:

Theorem: If $f:(a,b)\rightarrow\mathbb{R}$ is measurable, and $f$ is differentiable on a measurable set $C\subset(a,b)$, then $$m^*(f(C))\leq \int_C |f'(x)|\,dx$$ where $m$ is Lebesgue's measure and $m^*$ is the outer measure.

See Bruckner, A. M. et. al., Real Analysis, 2nd ed 2008, section 7.3. for example, or here

---

For your OP, consider $f:(0,1)\rightarrow(1,\infty)$ given by $f(x)=\frac{1}{x}$. Notice that $H=f(E)$. Applying the result stated above gives you $$ m^*(H)=m^*(f(E))\leq\int_E\frac{1}{x^2}\,dx=0 $$ since $m(E)=0$.

---

Edit: It is also possible to show directly without using big machinery that $m(H)=0$ using only the concept of measure $0$ and the fact that the countable union of sets of measure zero has also measure $0$ (Most Advanced Calculus or basic Analysis books discuss this, for example Apostol, T. Mathematical Analysis, 2nd ed. Section 7.26)).

I leave many details to the OP. Start by splitting $(0,1]$ in a countable subintervals, for example $\{(\tfrac{1}{n+1},\frac1n]:n\in\mathbb{N}\}$. The taks is then to show that each $H_n=\{\tfrac1x:x\in E\cap(\tfrac{1}{n+1},\frac{1}{n}]\}$ has measure zero. Notice that on each the function $f(x)=\frac{1}{x}$ is Lipschitz on $I_n:=[\frac{1}{n+1},\frac{1}{n}]$, There exists a constant $L_n$ such that $|f(x)-f(y)|\leq L_n|x-y|$ for all $x,y\in I_n$. Since $E\cap I_n$ has measure zero, then you can find a countable collection of intervals $J_{n,k}\cap _n=[a_{n,k},b_{n,k}$, $k\in\mathbb{N}$ such that $$\sum_k\ell(b_{n,k}-a_{n,k})<\frac{\varepsilon}{L_n}$$ where $\ell$ stands for length. Then $$\sum_k|f(b_{n,k})-f(a_{n,k})|\leq L_n\sum_k(b_{n,k}-a_{n,k})<\varepsilon$$ Notice that $f(J_{n,k}\cap I_n)$, $k\in\mathbb{N}$ is a collection of intervals that cover $H_n$. That shows that $H_n$ has measure zero.

Answer 2

For the record, here is the proof I had in mind when I wrote my comment. This is intended to address directly the problem implicit in the question, namely that taking reciprocals blows up the length of intervals near $0$. It uses two facts: (1) a set $E \subseteq \Bbb{R}$ has measure $0$ iff for any $\delta > 0$, there is a family of intervals $(x_n, y_n)$ , that cover $E$ (i.e., $E \subseteq \bigcup_i (x_i, y_i)$) and have total length less than $\delta$ (i.e., $\sum_i (y_i - x_i) < \delta$); (2) the union of a countable family of sets of measure $0$ has measure $0$.

Assume (1) holds for $E \subseteq (0, 1)$ and let $H_n = \{1/x \mid x \in E \cap [1/n, 1]\}$ for $n = 1, 2, \ldots$ Given $n$ and $\delta>0$, by our assumption, we can cover $E$ and hence $E \cap [1/n, 1]$ by intervals $(x_i, y_i)$ such that $\sum_i (y_i - x_i) < \delta/n^2$. But then $H_n$ is covered by the intervals $(1/y_i, 1/x_i)$ and we have: $$ \sum_i\left(\frac{1}{x_i}- \frac{1}{y_i}\right) = \sum_i\left(\frac{y_i - x_i}{x_iy_i}\right) < n^2 \sum_i(y_i - x_i) < \delta $$ because for $x_i, y_i \ge 1/n$, $1/(x_iy_i) < n^2$. So, by (1), each $H_n$ has measure $0$ and $H = \{1/x \mid x \in E\} = \bigcup_n H_n$ is a countable union of sets of measure $0$ and hence by (2) has measure $0$.