According to Terence Tao's Introduction to Measure Theory, a set $E \subset \mathbb R^d$ is said to be Lebesgue measurable if, for every $\epsilon > 0$, there exists an open set $U \subset \mathbb R^d$ containing $E$ such that $m^*(U\setminus E)\le \epsilon$. Using this definition, I was able to show that every set $E$ with $m^*(E)=0$ was measurable. Although the proof isn't very long, I wonder if there's a simpler way to see this than the arguments I used, as shown below:
Suppose $m^*(E)=0$. Then for any $\epsilon > 0$, there exists a countable union of boxes, $\bigcup_{n=1}^{\infty} B_n$, containing $E$, with $\sum_{n=1}^{\infty} |B_n|<\epsilon$. For each $B_n$, we then find an open box $C_n \supset B_n$ with volumn $|C_n| \le |B_n|+\frac{\epsilon}{2^n}$. As a result, $U \triangleq \bigcup_{n=1}^{\infty} C_n$ is an open set containing $E$ with $m^*(U)\le\sum_{n=1}^{\infty} |C_n|\le \sum_{n=1}^{\infty} \left(|B_n|+\frac{\epsilon}{2^n}\right)<2\epsilon$. Clearly, $m^*(U\setminus E) \le m^*(U)$, so $E$ is measurable.
Is this proof correct? Is there a more straightforward way to see it? Thanks a lot!
EDIT: This question is related to Are all measure zero sets measurable?, with the differences being (1) the definition of measurability is different, (2) this question is more general; it's about $\mathbb R^d$ instead of $\mathbb R$, (3) I want to verify a proof with these differences.
