Consider $E = \mathbb{R} \times \{0\}$ in $\mathbb{R}^2$. Apparaently this set (the x-axis) has measure zero. However, consider the following set
$$O_n = \big\{\mathbb{R} \times \big(-\frac{1}{n}, \frac{1}{n}\big)\big\} $$
and $\lim_{n \to \infty} m(O_n) = \lim_{n \to \infty} \frac{2}{n} \times \infty \neq 0$, where $m(*)$ is the Lebesgue measure.
Therefore, we don't have $\lim_{n \to \infty} m(O_n) = m(E)$.
Can you debunk my false proof of $\lim_{n \to \infty} m(O_n) = m(E)$?
Since $E$ is Lebesgue measurable, there is an open set $F$ such that $m(F\setminus E) < \epsilon$. Find $n$ such that $O_n \subset F$ and $m(O_n) \leq m(E) + \epsilon$, because $O_n$ can be made arbitrarily close to $E$. $\epsilon$ being arbitrary implies $m(O_n) \leq m(E)$ and $n \to \infty$ shows $\lim_{n \to \infty} m(O_n) \leq m(E)$.
The other direction $m(O_n) \geq m(E)$ is obvious since $O_n \supset E$ for all $n$.
What did I miss?
There is not necessarily any $n$ such that $O_n\subset F$, since $F$ might shrink arbitrarily close to $E$ as $x$ grows. For instance, $F$ might be $\{(x,y):|y|<1/(x^2+1)\}$.