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.

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    – ferhenk
    Commented Feb 6, 2021 at 12:58

1 Answer 1


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}


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