1.11

Notice that we can construct $\mathcal{A}$ in a similar fashion to how we construct $\mathcal{C}$. Proceed as follows. Define: $$\mathcal{A}_1 = [0, 0.4) \cup [0.5, 1].$$ $$\mathcal{A}_2 = \bigcup_{k=0}^{10^1 - 1} 10^{-1}([0 + k, 0.4 + k) \cup [0.5 + k, 1 + k])$$ $$\vdots$$ $$\mathcal{A}_n = \bigcup_{k=0}^{10^{n-1}-1} 10^{-n + 1}([0 + k, 0.4 + k) \cup [0.5 + k, 1+k])$$.

Now, just define $\mathcal{A} = \bigcap_{n=1}^{\infty} \mathcal{A_n}$. By corollary 3.3, since $\bigcap_{n=1}^{N} \mathcal{A_n} \searrow \mathcal{A}$ as $N \to \infty$, and $m(\mathcal{A_1}) = 0.9$, then: $$m(A) = \lim_{N \to \infty} m(\bigcap_{n=1}^{N} \mathcal{A_n}) = \lim_{N \to \infty} (0.9)^N = 0$$