On the Boundedness of Muckenhoupt $A_{1}$ Weights Near Infinity We say that a locally integrable $\omega(x)>0$ a.e. on $\mathbb{R}^{n}$ is an Muckenhoupt $A_{1}$ weight if ${\bf{M}}\omega(x)\leq C\cdot\omega(x)$ a.e. for some constant $C>0$, where ${\bf{M}}$ is the Hardy-Littlewood maximal function. And the infimum of all such $C>0$ is denoted by $[\omega]_{A_{1}}$.
Now my question is that, for $\omega\in A_{1}$, is necessarily that
\begin{align*}
\|\omega\|_{L^{\infty}(\{|x|>M\})}<\infty
\end{align*}
for some $M>0$?
Note that I am not asking if $\|\omega\|_{L^{\infty}(\mathbb{R}^{n})}<\infty$, for $\omega(x)=\log(1/|x|)$, $|x|<1/e$ and $\omega(x)=1$ otherwise, it is a standard fact that $\omega\in A_{1}$.
What I am asking is about the boundedness of such $\omega$ near infinity.
A well-known $A_{1}$ weight is given by $\omega(x)=1/|x|^{\eta}$ for $0\leq\eta<n$, but then this is bounded.
It seems that there are few examples of concrete $A_{1}$ weights, so I cannot come out with examples, any help is appreciated.
In fact, I am more interested in asking if the following is true for $\omega\in A_{1}$:
\begin{align*}
\sup_{z\in\mathbb{Z}^{n}}\left(\int_{B_{1}(z)}\omega(x)dx\right)<\infty?
\end{align*}
 A: All of those computations are very crude, I did not attempt to optimize constants. Feel free to edit if this bothers you.
Define $\omega_n(x) = \max\{ \frac{n}{\sqrt{\vert x \vert}}, 1 \}$ and
$$ \omega(x) = \begin{cases} \omega_n(x-4^n),& x\in [4^n-n^2, 4^n+n^2] \text{ for some } n\geq 3, \\ 1,& \text{otherwise}. \end{cases}$$
First we consider small radii. That means, $r$ is so small that it intersects at most one of the intervals $[4^n-n^2, 4^n+n^2]$. Then it is enough to check that $\omega_n$ is in $A_1$ (after a shift). We distinguish two cases. If $0< r< x_0 $
$$ \frac{1}{2r} \int_{x_0+r}^{x_0+r} \omega_n(x) \leq 1 +\frac{1}{2r} \int_{x_0+r}^{x_0+r} \frac{n}{\sqrt{\vert x \vert}}
= 1+\frac{n}{r} \left( \sqrt{x_0+r} - \sqrt{x_0-r} \right)
= 1+\frac{n}{r} \frac{2r}{\sqrt{x_0+r} + \sqrt{x_0-r}} \leq \frac{2n}{\sqrt{x_0}} \leq 3 \omega_n(x_0). $$
On the other hand, if $0<x_0\leq r$ (if $x_0=0$ it is trivial as $\omega_n(0)=\infty$ and by symmetry we can assume that $x_0>0$)
$$ \frac{1}{2r} \int_{x_0-r}^{x_0+r} \omega_n(x) \leq 1+\frac{1}{2r} \int_{x_0-r}^{x_0+r} \frac{n}{\sqrt{\vert x \vert}}
=1+ \frac{1}{2r} \left( \int_0^{r-x_0} \frac{n}{\sqrt{x}} + \int_{0}^{x_0+r} \frac{n}{\sqrt{x}} \right)
= 1+\frac{n}{r} \left( \sqrt{r-x_0} + \sqrt{x_0+r} \right)
\leq 1+\frac{n}{r} (1+\sqrt{2}) \sqrt{r} 
\leq (2+\sqrt{2}) \omega_n(x_0). $$
Finally, we consider large radii. Let $x\in [4^n, 4^{n+1}]$ and $4^m < r \leq 4^{m+1}$ for $m\geq n-1$. Then, we just integrate over all bumps that we could possibly reach
$$ \frac{1}{2r} \int_{x_0-r}^{x_0+r} \omega(x) \leq 1 + \frac{1}{2r} \sum_{j=1}^{m+1} 2 \int_0^{\sqrt{j}} \frac{j}{\sqrt{x}}
\leq 1 + \frac{2}{r} \sum_{j=1}^{m+1} j^2 \leq 1+ \frac{2}{4^m} \frac{(m+1) (m+2) (2m+3)}{6} \leq 4 \leq 4 \omega(x_0). $$
The remaining cases work similarly. Thus, we get that $\omega$ is indeed in $A_1$.
Note that this provides a counterexample as
$$ \int_{4^n}^{4^n+1} \omega(x) = \int_{4^n}^{4^n+1} \omega_n(x-4^n) = \int_0^1 \frac{n}{\sqrt{x}} = 2n. $$
