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
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.