For every open set $ G \subset \mathbb{R}^n$ and $ \epsilon > 0 $ there exist a finite collection of compact intervals $\{ I\_j \quad|\quad j=1,2,...m\}$ in $ \mathbb{R}^n $ such that $\cup^{j=1}\_{m}I\_j \subset G $ and $ m(G\backslash \cup^{j=1}\_{m}I\_j)< \epsilon $ where m denotes lebesgue measure.

NOTE : compact interval in $\mathbb{R}^n$ is $[a\_1,b\_1]\times[a\_2,b\_2]\times...\times[a\_n, b\_n]$ where $[a\_i,b\_i] $ are real closed intervals for each i.

**EDIT**: $m(G)$ is finite. sorry, I forget to write.
['https://math.stackexchange.com/questions/2840303', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/402639/']
This follows from the fact that every open set in $\mathbb{R}^n$ is a countable disjoint union of compact intervals. Indeed, if $G = \cup\_{k=1}^\infty I\_k$, then since $\sum\_k m(I\_k) = m(G) < \infty$, we may take $K$ so that $\sum\_{k > K} m(I\_k) < \epsilon$. Then $\cup\_{k=1}^K I\_k$ does the trick. 

See page 7 for the proof of the first fact I stated

<http://assets.press.princeton.edu/chapters/s8008.pdf>
This is false. For example take $G= \mathbb R^n$.