$A$ is a subset of $\mathbb{R}^2$ that for every $(x,y) \in A$ there is a $\delta >0$ that $(x-\delta , x+\delta) \times \{y\}$ and $\{x\} \times (y-\delta , y+\delta)$ are subsets of $A$. prove that $A$ is lebesgue measurable.
a set like $A$ may be not an open set, like the below set:
This is false. You can find a bijection from reals to reals whose graph $G$ is non measurable in plane. But the complement of $G$ is a $+$ set.
To construct such a function list all compact positive measure subsets of plane $\{K_{i} : i < 2^{\omega}\}$. Let $\{r_i : i < 2^{\omega}\}$ list all reals. Construct $f = \bigcup_{i < 2^{\omega}} f_i$ where
(1) $f_i$ is a partial injection from reals to reals with $|dom(f_i)| \leq max(|i|, \omega)$
(2) $\{r_j : j < i\}$ is contained in domain of $f_i$
(3) $\{r_j : j < i\}$ is contained in range of $f_i$
(4) The graph of $f_i$ meets $K_j$ for every $j < i$
At stage $i$, to satisfy (2), (3) simply add some appropriate pairs to $f$. To satisfy (4), note that for some $x$ not in the current domain of $f$, the vertical section of $K_i$ at $x$, $(K_i)_x = \{y : (x, y) \in K_i\}$ has size continuum. Pick some unused $y$ in here and add $(x, y)$ to $f$.