Exercise 1.7.1 (Null sets are Carathéodory measurable)

We use the usual trick of demonstrating that the left-hand side is both less than or equal to and greater than or equal to the right-hand side.

• We have $A=(A\cap E)\cup (A\setminus E)$ and thus, by subadditivity of the outer measure

  $${\mu }^{\text{∗}}(A)\le {\mu }^{\text{∗}}(A\cap E)+{\mu }^{\text{∗}}(A\setminus E).$$

• Notice that $A\cap E\subseteq E$ and therefore, by monotonicity of the outer measure, ${\mu }^{\text{∗}}(A\cap E)\le {\mu }^{\text{∗}}(E)=0$. By non-negativity of the outer measure we conclude that ${\mu }^{\text{∗}}(A\cap E)=0$. By the monotonicity again, we obtain

  $${\mu }^{\text{∗}}(A)\ge {\mu }^{\text{∗}}(A\setminus E)$$

  as desired.