Show that no non-trivial open set in $\mathbb{R}^n$ has measure zero in $\mathbb{R}^n$.
What does non-trivial mean? (I think it means that the set contains a general subset of $\mathbb{R}^n$, doesn't it?)
Any idea for a rigorous proof?
My proof:
Let $A$ be the non-trivial open set in $\mathbb{R}^n$. It contains a closed rectangle $Q$. Since $A$ is open, there exist an open covering of $Q$, say {$Q_i$}$^{\infty}$, contained in $A$; moreover a finite subcollection, say {$Q_1,…,Q_n$}, covers $Q$ (choose the open covering so that each set has at most boundary points in common). Thus:$$\sum_{i=1}^{n}v(Q_i)\ge v(Q)$$
where $v(Q)$ is the volume of $Q$.
Now let {$A_i$}$^{\infty}$ be a countable collection of rectangles which covers $A$. My attempt is to show that the total volume of the rectangles $A_1,A_2,…$ cannot be made less than $v(Q)$; since {$A_1,A_2,…$} covers $A$, it will also covers {$Q_1,…,Q_n$}; therefore:$$\sum_{i=1}^{\infty}v(A_i)\ge \sum_{i=1}^{n}v(Q_i)$$
so that one ends the proof.
Is this correct?
Best Answer
if A is non-empty (which is what non-trivial means), it contains a point $x$. So there exists $r >0,$ such that $$ B(x,r) $$ is contained in $A$ since $A$ is open.
The measure of S is bigger than that of $B(x,r)$ which has the measure of a ball of radius $r$ in $R^n$ which is positive.