Let $m^*$ be the Lebesgue outer measure in $\mathbb{R}$ and suppose $E \subset \mathbb{R}$ satisfies its outer measure is finite.
Prove that for each $\varepsilon > 0$, there exists $M>0$ such that $m^* (E \setminus [-M,M]) < \varepsilon$.
I have a problem here. Because we don't know whether $E$ is measurable or not, so I cannot apply the continuity of measure or the inner/outer approximation to do.
All the things I have is the definition of outer measure and since $[-M,M]$ is measurable,
$$m^* (E) = m^* (E \cap [-M,M]) + m^* (E \setminus [-M,M]).$$
The textbook I used is Real Analysis by Royden and Fitzpatrick.
Any help is highly appreciated.
Take $\{B_{j}\}$ boxes so that $E\subseteq \bigcup B_{j}$ and $\sum |B_{j}|\leq m^{*}(E)+ 1$. Because $m^{*}(E)$ is finite, there is a big enough $N$ so that $\sum\limits_{j \geq N}|B_{j}|\leq \varepsilon$. The family $\{B_{j}\}_{j\leq N}$ is bounded, so there is a big enough $M$ so that $$\bigcup\limits_{j\leq N}{B_{j}}\subseteq [-M,M]$$ We have that $\{B_{j}\}_{j \geq N}$ covers $E\setminus [-M,M]$ so $$m^{*}(E\setminus [-M,M])\leq \varepsilon$$