Measure of difference of sets

Question

Suppose $G$ is an open (measurable) set, then $G^{c}$ is also measurable. Then, I did this, is it right?: \begin{align*} \lambda(G^{c} \cap [0,1])&=\lambda(G^{c}-(G^{c}-[0,1]))\\ &=\lambda(G^{c})-\lambda(G^{c})+\lambda([0,1])\\ &=\lambda([0,1])=1 \end{align*}

Answer 1

There are a few ways to see this is wrong. The first is a simple counterexample: take $G = \mathbb R$ in which case $G^c = \varnothing = G^c \cap [0, 1]$ so $\lambda(G^c \cap [0, 1]) = 0$.

The second is if $\lambda(G^c \cap [0, 1]) = 1$ then $\lambda((G^c)^c \cap [0, 1]) = \lambda(G \cap [0, 1]) = 1$ by the same reasoning. But $G^c \cap [0, 1]$ and $G \cap [0, 1]$ are disjoint so by countable additivity of Lebesgue measure $$ \lambda((G^c \cap [0, 1]) \cup (G \cap [0, 1])) = 2 $$ but this is a contradiction since $(G^c \cap [0, 1]) \cup (G \cap [0, 1]) = [0, 1]$.

Another is that were this argument correct, then $\lambda(G^c) = \infty$ for all $G$ because you can replace $[0, 1]$ with $[-n, n]$ and take $n \rightarrow \infty$.

The issue is that $\lambda(A - B) = \lambda(A \cap B^c) = \lambda(A) - \lambda(B)$ only when $B \subseteq A$.