As part of a proof the following was written:
For any Lebesgue measurable set $D$ there exists a $G_{\delta}$ set, $A$,such that $D \subseteq A$ and $m(A\setminus D)=0$. If we define the following indicator functions: $I_D$ and $I_A$ are one if the point is in $D$ (or $A$) and zero if the point is not in $D$ (or $A$), then $I_A\neq I_D$.
a. Why are they not equal? I think they mean not equal almost everywhere
b. How can $D\subseteq A$? Isn't a Lebesgue set "bigger" than Borel set?
It is a standard theorem in measure theory that a set $E$ is Lebesgue measurable if and only if there exists a $G_\delta$ set $G$, such that $G\supset E$ and $m(G\backslash E)=0$, and a $F_\delta$ set $F$, such that $F\subset E$ and $m(E\backslash F)=0$.
a) Now I'm not sure what proof you're referring to, but in general we aren't guaranteed that $\mathbb 1_A\neq\mathbb 1_D$. Just consider the case where $D=\emptyset$, then setting $A=\emptyset$ satisfies all the requirements, so their indicator functions are identical. You will have to give more detail to exactly what the proof is for and what the nature of $D$ is.
b) The Lebesgue $\sigma$-algebra is "bigger" than the Borel $\sigma$-algebra. By this we mean that if a set is Borel measurable then it is also Lebesgue measurable, but there are Lebesgue measurable sets that aren't Borel measurable. This says nothing about the "size" of individual sets. For example $\mathbb R$ is Borel measurable and $\emptyset$ is Lebesgue measurable, but quite clearly $\emptyset\subset \mathbb R$.