Question

Question 1: In $\mathbb{R}^2$, let $l_1$,$l_2$ be two parallel lines and $l_3$ another line which is not parallel to $l_1$. Given two measurable sets $E_1$ and $E_2$ in $l_1$ and $l_2$ respectively, both of which have positive (1-dimension) Lebesgue measure. Denote by $l_{u,v}$ the line passing through two points $u,v$. Now, we define the third set $E_3$ in $l_3$ as follows. $E_3=\lbrace{p\in l_3: p=l_{u,v}\cap l_3, u\in E_1, v\in E_2\rbrace}$.

Can we say that $E_3$ contains an interval (or say "a segment") in $l_3$?

EDIT. Joseph Van Name answered Question 1 affirmatively. The following Question 2 is a generalization of Question 1, which is not answered.

Question 2: In $\mathbb{R}^n$, let $L_1$,$L_2$ be two parallel (n-1)-dimensional hyperplanes and $L_3$ another (n-1)-hyperplane which is not parallel to $L_1$. Given two measurable sets $E_1$ and $E_2$ in $L_1$ and $L_2$ respectively, both of which have positive ((n-1)-dimension) Lebesgue measure. Denote by $l_{u,v}$ the line passing through two points $u,v$. Now, we define the third set $E_3$ in $L_3$ as follows. $E_3=\lbrace{p\in L_3: p=l_{u,v}\cap L_3, u\in E_1, v\in E_2\rbrace}$.

Can we say that $E_3$ contains an $(n-1)$-ball in $L_3$?

It seems that the result in "An elementary proof and an extension of a theorem of Steinhaus,Kuczma, Marcin E.; Kuczma, Marek, Glasnik Mat. Ser. III 6(26) (1971), 11–18." noted by Joseph Van Name does not apply to the general case $n\geq 3$ of Question 2.

Answer 1

We will need the following result from the paper that I mentioned in the comments "An elementary proof and an extension of a theorem of Steinhaus" by M. Kuczma and M. Kuczma that generalizes the Steinhaus theorem.

Thm:(Kuczma and Kuczma) Let $A,B$ be sets with positive inner Lebesgue measure and let $f(x,y)$ be a real-valued function of class $C^{1}$ in a region $D\supseteq A\times B$ with $f_{x}\neq 0,f_{y}\neq 0$ in $D$. Then the set $f(A\times B)$ contains an interval.

Now, to prove our result we shall assume without loss of generality that $\ell_{1}=\{(x,0)|x\in\mathbb{R}\},\ell_{2}=\{(x,1)|x\in\mathbb{R}\}$ and $\ell_{3}=\{(0,y)|x\in\mathbb{R}\}$. Let $U=\{(x,y)\in\mathbb{R}^{2}|x\neq y\}$. Then $U$ is the union of the two regions $\{(x,y)|x<y\}$ and $\{(x,y)|x>y\}.$ Let $f:U\rightarrow\mathbb{R}$ be the function where if $(x,y)\in U$, then $f(x,y)$ is the unique real number such that the points $(x,0),(y,1),(0,f(x,y))$ all fall on the same line. Then any reasonable high school algebra student should be able to tell you that $f(x,y)=\frac{x}{x-y}$. Therefore $f$ is real-analytic on $U$. Also, any calculus 3 student should also be able to tell you that $f_{x}(x,y)=\frac{-y}{(x-y)^{2}}$ and $f_{y}(x,y)=\frac{x}{(x-y)^{2}}$. Therefore the first partial derivatives of $f$ with respect to $x$ and $y$ do not vanish on the set $U$.

If $E_{1},E_{2}\subseteq\mathbb{R}$ are sets with positive Lebesgue measure, then there are subsets $A_{1}\subseteq E_{1},A_{2}\subseteq E_{2}$ of positive Lebesgue measure such that either $a_{1}<a_{2}$ for $a_{1}\in A_{1},a_{2}\in A_{2}$ or $a_{1}>a_{2}$ for $a_{1}\in A_{1},a_{2}\in A_{2}$. Therefore the set $A_{1}\times A_{2}$ is completely contained in some region in $U$, so $f(A_{1}\times A_{2})$ contains some interval. We therefore conclude that the set $E_{3}$ defined in the problem must contain some interval as well.