The Lebesgue measure of the set of horizontal lines through the points of a subset $A$ of $\mathbb{R}$ with $\lambda(A)=0$ Suppose $A$ is a subset of the real line with $\lambda(A)=0$ and $H=\{(x,y):x\in A\}$. What is a natural idea behind proving that $\lambda(H)=0$ ? In fact, I wish to prove that the collection of horizontal lines through the points of a Lebesgue measurable subset of the real line is a Lebesgue measurable subset of the plane,and one way of proving it is to show that it's a union of an $F_\sigma$ and a set with Lebesgue measure zero. Thanks in advance
 A: \begin{align*}
\lambda (H) &= \int _{\mathbb{R}} \int _{\mathbb{R}} 1_{H  }(x,y)dx dy\\
&=\int _{\mathbb{R}} \lambda (A) dy  \\
&=0
\end{align*}
A: Let $H_n = A \times [-n,n]$. Then $H_n$ is measurable and $\lambda_2(H_n) = 0$ by Fubini's theorem.  Since $H = \cup_n H_n$ it follows that $H$ is measurable too, and the subadditivity of the measure gives you $\lambda_2(H) = 0$.

Added on edit: fix some index $n$ and let $\epsilon > 0$ be some positive number. Since $\lambda(A) = 0$ there exists, by definition, a countable family of open intervals $\{I_k\}$ with the property that $$A \subset \bigcup_k I_k \quad \text{and} \quad  \sum_k \ell(I_k) < \epsilon.$$
Now define a family $R_k$ of open rectangles by $R_k = I_k \times (-2n,2n)$. Then $$H_n \subset \bigcup_k R_k$$ which implies $$\lambda_2^*(H_n) \le \sum_k A(R_k) = \sum_k \ell(I_k) \cdot 4n < 4n\epsilon$$ where the star denotes the outer measure.
Since this is valid for any $\epsilon > 0$ this forces $\lambda_2^*(H_n) = 0$ which implies that $H_n$ is measurable. From here you get $H$ measurable and $\lambda(H) = 0$ as above.
