Every open set $ G \subset \mathbb{R}^n$ can be approximated by finite union of non-degenerate compact intervals contained in $G$? For every open set $ G \subset \mathbb{R}^n$ and $ \epsilon &gt; 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)&lt; \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.
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 $\sum_k m(I_k) = m(G) &lt; \infty$, we may take $K$ so that $\sum_{k &gt; K} m(I_k) &lt; \epsilon$. 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 $G= \mathbb R^n$.