# Show that no non-trivial open set in $\mathbb{R}^n$ has measure zero in $\mathbb{R}^n$

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?

• non-trivial just means non-empty in this context Apr 4 '15 at 12:38
• @user228695 $A$ isn't open in $\Bbb R^2$. Apr 4 '15 at 12:42
• Actually, you can almost stop right after "It contains a closed rectangle $Q$", for $\mu(A)\ge \mu(Q)>0$ (if $A$ is measurable at all). Apr 4 '15 at 12:49
• What do you mean? Apr 4 '15 at 12:50
• Hagen, was my assertion correct? Apr 4 '15 at 12:56

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.
Let $$U$$ be a non-empty open set in $$\mathbb{R}^n$$.
Assume that $$U$$ has measure zero in $$\mathbb{R}^n$$.
Since $$U$$ is open, there is a (closed) rectangle $$Q$$ which is contained in $$U$$.
Since $$U$$ has measure zero in $$\mathbb{R}^n$$, for $$\epsilon=v(Q)$$, there is a covering $$\text{Int}\,Q_1,\text{Int}\,Q_2,\dots$$ of $$U$$ by countably many open rectangles such that $$\sum_{i=1}^{\infty}v(Q_i)<\epsilon. \text{(See Theorem 11.1(c) on p.91.)}$$ Since $$Q$$ is compact, there is a finite subcollection $$\text{Int}\,Q_{i_1},\dots,\text{Int}\,Q_{i_k}$$ that covers $$Q$$.
So, the finite collection of (closed) rectangles $$Q_{i_1},\dots,Q_{i_k}$$ covers $$Q$$.
By Corollary 10.5 on p.88, $$v(Q)\leq\sum_{j=1}^{k}v(Q_{i_j})\leq\sum_{i=1}^{\infty}v(Q_i)<\epsilon=v(Q).$$ This is a contradiction.