Open sets and measure zero

Question

Show that no non-trivial open set in $R^n$ can have measure zero in $R^n$.

Attempt at the solution: I am having a lot of difficulty attempting this question, I have read a lot of material on measure zero and almost all of the questions posted on this forum regarding the topic, but I still can't seem to understand how to attempt this problem. I have written a proof that somewhat makes intuitive set to me, but I am pretty terrible at proof writing and trying to improve so please don't be sparing in your recommendations and criticism.

Let A be open set in $R^n$ of the form (a,b) s.t a $\not=$ b, we can choose intervals I for any $x_i$ $\in$ A :[$x_i - \epsilon $, $x_i + \epsilon$], we can make these intervals arbitrarily small, the problem occurs at the points a and b, we will need an interval that covers a and closest point to a that is included in the set, the smallest interval of this form would be [a-$\epsilon$, $x_1 + \epsilon$], this interval does not have measure zero by the definition of open sets ( since if a is arbitrarily close to $x_1$ we would arrive at a contradiction i.e A will no longer be open), which gives us that no non trivial open set can have measure zero. Thanks in advance

The open sets in $\mathbb{R}^n$ are not intervals, they are generated by open balls. Therefore, your first step in the proof needs to be changed to use an open ball. — Michael Burr, Dec 26 '15 at 17:47

score 4 · Accepted Answer · answered Dec 26 '15 at 17:52

4

Your proof has the right idea, but could be made cleaner. If I understand your approach, for a fixed $x$, you are constructing a closed $n$-square around $x$ that is contained in $A$. Then, since the measure of $A$ is greater than or equal to the measure of the $n$-square, which is nonzero, the measure of $A$ is not zero.

Hint: It may be easier to start with $x\in A$. Since $x\in A$ and $A$ is open, there is some $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq A$. Now, it is easy to construct a closed $n$-square within the open ball, for instance there is an $n$-square with side length $\frac{\sqrt{2}\varepsilon}{\sqrt{n}}$ centered at $x$ which is contained within the ball.

answered Dec 26 '15 at 17:52

Michael Burr

33,866

If could even be $\frac{2\varepsilon}{\sqrt n}$. – hmakholm left over Monica Dec 26 '15 at 18:08
@HenningMakholm If you use an $n$-square of that size, then the farthest corner will be at distance $\varepsilon$ from $x$ and the OP mentioned a closed $n$-square. Anything less than $2$ in the numerator will work, however for a closed $n$-square. – Michael Burr Dec 26 '15 at 18:22
1

I see. (Though the open hypercube has the same measure, of course). – hmakholm left over Monica Dec 26 '15 at 18:33
hey, thank you for your answer, how did you come up with the side length ? – ಠ_ಠ Dec 28 '15 at 22:57
I took the diagonal of the box and divided by $2$. – Michael Burr Dec 28 '15 at 23:29
Isn't the longest diagonal $\sqrt{n}$ , I understand why you said anything less than $2\epsilon$ but why divide by the diagonal – ಠ_ಠ Dec 28 '15 at 23:53
Can we reach the same conclusion as above by using a square of side $\sqrt{2}\epsilon$ – ಠ_ಠ Dec 29 '15 at 01:40
For an $n$-cube with side length $2s$, the corner is distance $\sqrt{n}s$ from the center. To be within a ball of radius $\varepsilon$, it must be that $s<\frac{\varepsilon}{\sqrt{n}}$. – Michael Burr Dec 29 '15 at 03:26

Open sets and measure zero

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