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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?

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    $\begingroup$ non-trivial just means non-empty in this context $\endgroup$
    – Rolf Hoyer
    Apr 4 '15 at 12:38
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    $\begingroup$ @user228695 $A$ isn't open in $\Bbb R^2$. $\endgroup$ Apr 4 '15 at 12:42
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    $\begingroup$ 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). $\endgroup$ Apr 4 '15 at 12:49
  • $\begingroup$ What do you mean? $\endgroup$
    – user228695
    Apr 4 '15 at 12:50
  • $\begingroup$ Hagen, was my assertion correct? $\endgroup$
    – user228695
    Apr 4 '15 at 12:56
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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.

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I guess this exercise is from "Analysis on Manifolds" by James R. Munkres. (Exercise 2. on p.97.)

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.

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