Is there a constant $A$ such that the measure of the set $x\in [0,1]$ such that $|x-p/q|>(Aq^2)^{-1}$ for any rational $p/q$ is equal to 1?
While trying to find the answer to this question, I found the following mathoverflow post: https://mathoverflow.net/questions/105480/are-there-lower-bounds-on-the-quality-of-a-rational-approximation
In particular, the answer in this post seems very relevant to my question. In the 3rd bullet point, it is stated there is continuum many $x$ such that $|x-p/q|>(3q^2)^{-1}$ for any rational $p/q$. I don't know if this would imply an affirmative answer to my question with $A=3$, I am not well versed in number theory. If not, is my question true for some other $A$?
Definitely not. The complement of your set is the union of the countably many closed intervals $$ \bigg[ \frac pq-\frac1{Aq^2},\frac pq+\frac1{Aq^2} \bigg], $$ and therefore the measure of the complement is at least as large as the measure of any one of those closed intervals—so, for example, is at least $\frac1{2A}$ (assuming $A\ge\frac12$) by considering the interval around $\frac pq=\frac12$.