Edit: In writing out more details, I realized I needed to make a slight modification to the question. I switched all proper containments from the old version of the question to "essential containment".
This question arises in looking for a counterexample to a claim made in the book by Dunford and Schwartz, but seems interesting in its own right. The question I really want to answer is: "Does there exist a Lebesgue measurable set $C \subseteq [0,1]$ with $m(C)>0$ s.t. for every Lebesgue measurable $m(S \cap C^c) = 0$ s.t. $m(S) >0$, the set $\{x : m((x+S) \cap C^c) =0\}$ is Lebesgue null?"
It would suffice to show the following claim which seems possibly true: "Let $C$ a fat Cantor set. Then for every $S \subseteq C$ s.t. $m(S)>0$, the set $\{x : m((x +S) \cap C^c)=0\}$ is countable."
Edit: Due to a request in the comments, I'll post a simplification of the claim made in Dunford and Schwartz that inspired the question. Let $(A, \mathcal{A},\alpha), (B, \mathcal{B},\beta)$ $\sigma$-finite measure spaces and $(A \times B, \mathcal{A} \otimes \mathcal{B}, \alpha \times \beta)$ their product. Let $f : A \times B \to \mathbb{R}$ measurable. Then there exists a sequence $\phi_n$ of finite linear combinations of indicators on measurable rectangles (sets of the form $S \times T, S \in \mathcal{A}, T \in \mathcal{B}$) s.t. $\phi_n \to f$ pointwise a.e. and $|\phi_n| \leq |f|$.
If we have a set $C$ of the above type, let $D := \{(x,x) + (c,0) : c \in C, 0 \leq x \leq 1\}.$ Then the above property shows that if $\phi$ is a finite linear combination of indicators of measurable rectangles s.t. $|\phi| \leq 1_D$, then $\phi = 0$ a.e. This contradicts the Dunford and Schwartz claim as we can show that $m(D)>0$.
