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Let $\mu^{*}$ be the Lebesgue outer measure on $\mathbb{R}$. I found the following exercise in a textbook.

Exercise. For every $A \subseteq \mathbb{R}$, $\lim_{k \to \infty} \mu^{*}(A \cap [-k,k]) = \mu^{*}(A)$.

I can get the case $\mu^{*}(A) < \infty$, but I am stuck with the case $\mu^{*}(A) = \infty$. Please help. Not homework, just reading.

Since the exercise appears before the introduction of measures, I presume that the countable additivity and continuity of the Lebesgue measure should not be used. This is in contrast to the answers: https://math.stackexchange.com/a/1577811/95800 https://math.stackexchange.com/a/117252/95800

Here is my argument for the case $\mu^{*}(A) < \infty$. There exists a sequence of intervals $I_1,I_2,\ldots$ such that $A \subseteq \bigcup_{i=1}^{\infty} I_i$ and $\sum_{i=1}^{\infty} \ell(I_i) < \infty$. Let $\epsilon > 0$ be given. There exists $N \in \mathbb{N}$ such that $\sum_{i>N} \ell(I_i) < \epsilon$. Choose $K \in \mathbb{N}$ such that $\bigcup_{i=1}^{N} I_i \subseteq [-K,K]$. Then $A \setminus [-K,K] \subseteq \bigcup_{i > N} I_i$. Moreover, $A \setminus [-k,k] \subseteq \bigcup_{i > N} I_i$ for every $k \geq K$. Therefore $$ \mu^{*}(A) - \mu^{*}(A \cap [-k,k]) \leq \mu^{*}(A\setminus [-k,k]) \leq \sum_{i>N} \ell(I_i) < \epsilon $$ for every $k \geq K$. This proves $\mu^{*}(A) = \lim_{k \to \infty} \mu^{*}(A \cap [-k,k])$.

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