Dirichlet's theorem says that all numbers $x\in [0,1]$ can be approximated by infinitely many fractions $p/q \in \mathbb{Q}$ with error $|x - p/q| \le 1/q^2$.


I am interested in the following question:
Fix $m>0$. Let $S(m)$ denote the set of $x\in [0,1]$ for which no fraction $p/q \in \mathbb{Q}$ with $q<100\sqrt{m}$ approximates $x$ with error smaller than $1/m$, i.e., $|x-p/q| > 1/m$ for any such choice of $p,q$.
What bounds are there for the measure of $S(m)$?


EDIT:
Let $\Phi(q)$ denote the set of numbers smaller than $q$ coprime to $q$.
And let
$$I(q) = \bigcup\_{p\in \Phi(q)} [p/q, p/q+1/m]$$
For $q\_1,q\_2<\sqrt{m}$ the sets $I(q\_1),I(q\_2)$ will be disjoint.
This can give a simple lower bound (by analyzing $\sum\_{k<\sqrt{m}} \phi(k)$, which is a well known problem). I'm interested in if anything stronger than this bound can be given.



Simultaneously approximating all $x \in [0,1]$ with fractions of bounded denominator
Fix $T$ ($T=100$ in your example) and denote by $S\_T(n)$ the set of numbers $x\in [0,1]$ such that $|x-p/q|>1/n^2$ whenever $p,q$ are integers and $1\leqslant q\leqslant Tn$ (so, my $n$ is your $\sqrt{m}$).


Let $0=r\_0<r\_1<\ldots<r\_K=1$ be Farey sequence of all irreducible fractions with denominator at most $Tm$. Denote $\delta\_i=r\_{i+1}-r\_i$ for $i=0,1,\ldots,K-1$ the gap between corresponding Farey fractions. Then $S\_T(n)\cap [r\_{i},r\_{i+1}]$ has length $\max(\delta\_i-2/n^2,0)$, so, the measure of $S\_T(n)$ equals
$$
f(T,n):=\sum\_{i:\delta\_i>2/n^2} (\delta\_i-2/n^2).
$$
Recall that if $r\_i=a/b$, $r\_{i+i}=c/d$ are irreducible fractions, then $\delta\_i=1/(bd)$ and $b+d>Tn$. Denote $M=\max(b,d)$, $m=\min(b,d)$. Then $M> Tn/2$, and the inequality $1/(Mm)=\delta\_i>2/n^2$ yields $1/m>2M/n^2>T/n$, so, $m<n/T$. And $\delta\_i=1/(mM)\leqslant 2/(Tnm)$. Thus, for fixed $m$, the fractions with denominator $m$ give a contribution at most
$$
2\varphi(m)\left(\frac{2}{Tnm}-\frac2{n^2}\right)\leqslant \frac4{Tn}.
$$
Summing up over $m<n/T$ gives you at most $4/T^2$.


On the other hand, if $m<n/(3T)$, say, then $mM\leqslant mTn\leqslant n^2/3$ and thus $1/bd\geqslant 3/n^2$. Therefore, each of these intervals (with $m<n/(3T)$) give contribution at least $1/n^2$, and there are about $$2\sum\_{m<n/(3T)} \varphi(m)\sim \zeta(2)\frac{n^2}{9T^2}$$
such intervals that gives lower bound also of order $1/T^2$.


So, $$c\_1/T^2<f(T,n)<c\_2/T^2$$
for some universal explicit constants $c\_1,c\_2$. If you care on sharp constant, please let me know, this requires more accurate analysis but looks doable.


