Let S ⊆ J, where J is a closed n-dimensional interval, and let f: J → R be defined by
$$ f(x) = \begin{cases} 1 & x\in S \\\\ 0 & x ∈ J\backslash S. ~ \end{cases} $$
Let A be the set of all the interior points of $S$ and let B be the set of all the interior points of $S^c$ . Show that $$\int_{J}^{}f$$ exists if and only if $$λ ((A ∪ B)^c ∩ J) = 0$$
Only the implication $\Rightarrow$ of your statement is true, and we have also to suppose that the measure on $\mathbb{R}$ is complete, i.e. $\lambda$ is the Lebesgue measure (the completion of the Borel measure).
Since $J$ is an $n$-dimensional interval, it is bounded. Thus, we have only to prove that the set $S$ is measurable if and only if $\lambda((A\cup B)^c \cap J) =0$. Note that $A$ and $B$ are open and $S \setminus A \subset B^c$. Thus $(S \setminus A) \cap B^c = S \cap (A \cup B)^c \subset J \cap (A\cup B)^c$. Now, if $\lambda((A\cup B)^c \cap J) =0$, then is $S \setminus A$ is a nullset and thus Lebesgue measurable. Hence $S= A \cup S \setminus A$ is measurable.
One counterexample was already stated: $S= \mathbb{Q} \cap J$ is measurable with $\lambda(S) =0$, but $A= \emptyset$ and $B = \emptyset$. Hence, $(A \cup B)^c \cap J = J$ and $\lambda(J) > 0$.