Let $\mu$ be a Borel probability measure on $\mathbb{R}^n$ such that there are $c,C,d,D>0$ satisfying: for every $x \in \mathbb{R}^n$ and every $r>0$
$$
c r^d \leq \mu(B(x,r)) \leq Cr^D.
$$
Let's call such a measure **almost Ahlors Regular.**


***Note:** when $d=D$ then we say that the measure is [*Ahlfors regular*](https://mathoverflow.net/questions/319955/origin-of-term-ahlfors-david-regular) and, if $c$ were allowed to be $=0$ then the measure would be called upper [*Ahlfors regular*](https://mathoverflow.net/questions/415903/terminology-upper-ahlfors-regular-measure).*


What are examples of Ahlors regular measures?



Examples of "almost" Ahlfors regular measures
$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$A measure $\mu$ is Ahlfors regular (according to your definition) iff it has a density (with respect to the Lebesgue measure $\la$) bounded away from $0$ and $\infty$ (if $B(x,r)$ denotes the ball centered at $x$ and of radius $r$).


Indeed, by inscribing and circumscribing, we can replace $B(x,r)$ in your definition by the (left-open) $n$-cube $b(x,r):=\prod\_1^n(x\_i,x\_i+r]$ (say of size $r$) for $x=(x\_1,\dots,x\_n)\in\R^n$ (at the same time also appropriately changing the values of $c$ and $C$), so that
$$
c r^d\le\mu(b(x,r))\le Cr^d \tag{1}\label{1}
$$
for all $x\in\R^n$ and $r>0$.


Partitioning, in the natural manner, the unit $n$-cube $b(0,1)$ into $N^n$ $\ n$-cubes each of size $1/N$, from \eqref{1} we get
$$
c N^{n-d}\le\mu(b(0,1))\le CN^{n-d} \tag{2}\label{2}
$$
for all $N$, which is possible only when $d=n$ -- otherwise, $N^{n-d}$ will go either to $0$ or to $\infty$ as $N\to\infty$, which would contradict one of the two inequalities in \eqref{2}, since $\mu(b(0,1))\in(0,\infty)$.


Therefore and by partitioning and approximation, we get
$$
c \prod\_1^n r\_i\le\mu\Big(\prod\_1^n(x\_i,x\_i+r\_i]\Big)\le C \prod\_1^n r\_i 
$$
for any real $x\_1,\dots,x\_n$ and any positive real $r\_1,\dots,r\_n$.
That is, $c\la\le\mu\le C\la$ on the semiring of all left-open $n$-boxes.


We conclude that $c\la\le\mu\le C\la$ on the Lebesgue $\sigma$-algebra, as claimed.


