Show $|A \cap B_r(0)^c| \to 0$ for any set $A$ with finite Lebesgue measure

Question

Let $A \subset R^n$ satisfy $|A| < \infty$ then $|A \cap B_r(0)^C| \rightarrow 0$ as $r \rightarrow\infty$.

Don't know how to start. For example we can write

$$B_r(0)^C = \bigcup_{k=r+1}^\infty B_k \backslash B_{k-1}$$

Answer 1

Hint: (Here, $\lambda$ denotes the Lebesgue measure.)

1. The sequence $$C_r := A \cap B_r(0)^c$$ is decreasing and $$\bigcap_{r \in \mathbb{N}} C_r = \emptyset.$$
2. The finite measure $\lambda|_A$ is continuous from above, i.e. $$\lim_{r \to \infty} \lambda(C_r) = \lambda(C)$$ for any decreasing sequence $C_r$ of Borel sets such that $\bigcap_{r \in \mathbb{N}} C_r = C$.

Remark: If you know the dominated convergence theorem, then the statement follows also easily from this theorem.

Answer 2

For $k \ge 1$ integer, note $A_k = A \cap B_k(0)$. $(A_k)$ is an increasing sequence of subsets of $A$ and $$A=\bigcup_{k \ge 1}A_k$$ Hence for $\epsilon >0$, you can find $k_\epsilon \in \mathbb N$ such that $\mu(A) \ge \mu(A_k) > \mu(A) (1-\epsilon)$ for $k \ge k_\epsilon$.

For $r \ge k_\epsilon$ you also have $0 \le \mu(A \cap B_r(0)^c) \le \mu(A)\epsilon$, which concludes the proof.

$\mu $ denotes Lebesgue measure.