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?

  • 4
    $\begingroup$ non-trivial just means non-empty in this context $\endgroup$
    – Rolf Hoyer
    Apr 4 '15 at 12:38
  • 1
    $\begingroup$ @user228695 $A$ isn't open in $\Bbb R^2$. $\endgroup$ Apr 4 '15 at 12:42
  • 2
    $\begingroup$ 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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.