# Prove that $m_{*}(A)=\lim_{n\rightarrow \infty}m_{*}(A\cap (-n,n))$

How does one go about proving $$m_{*}(A)=\lim_{n\rightarrow \infty}m_{*}(A\cap (-n,n))$$ where $$m_{*}$$ is the outer measure? Here $$n\in \mathbb{N}$$.

I was considering the notation

$$A=\bigcup_{n=1}^{\infty}A_n$$ where $$A_n=A\cap(-n,n)$$ along with the property that

$$m_{*}\Big(\bigcup_{n=1}^{\infty}A_{n} \Big) \le 2\sum_{n=1}^{\infty}n$$

since $$m_{*}(A\cap (-n,n))\le m_{*}((-n,n)) = 2n$$. Here I get stuck and I am not sure what to do.

• Please make sure to accept responses if they answer your question. There are many more answers to your questions in your profile which are not confirmed. Feb 6, 2021 at 12:58

So, I assume we are considering an outer measure $$m_*$$ on $$\mathbb{R}$$. Set $$S_n := (-n,n)$$. Then, $$S_n \uparrow \mathbb{R}$$ as sets. By the definition of monotonicity, it suffices to prove that $${m_*} (A) \le \lim_{n \mathop \to \infty} {m_*} (A \cap S_n)$$ Assume that $$m_*({A \cap S_n})$$ is finite for all $$n \in N$$, otherwise the statement is trivial by the monotonicity of $$m_*$$.
Clearly, $$\mathbb{R} = \bigcup_{n} S_{n+1}\setminus S_n$$ and hence $$A = \bigcup_{n} A \cap (S_{n+1} \setminus S_n)$$. Assuming that $$S_n$$ is $$m_*$$-measurable, this yields \begin{align} m_*(A) &\leq \sum_{n}m_*(A \cap (S_{n+1} \setminus S_n)) \\ & =\sum_{n} m_*(A \cap S_{n+1}) - m_*(A \cap S_{n+1} \cap S_{n})\\ &=\sum_{n} m_{*}(A \cap S_{n+1}) - m_*(A \cap S_n)\\ &= \lim_{n \to \infty}m_*(A \cap S_n) - m_*(A \cap \emptyset)\\ &= \lim_{n\to \infty} m_*(A \cap S_n) \end{align}