This is a follow-up to [this question](https://mathoverflow.net/questions/392820/). We say that a set $A \subseteq \mathbb{R}$ is bounded if there exists a finite interval $(a,b)$ such that $A \subseteq (a,b)$.

Working in $\mathsf{ZFC}$, the existence of a (Lebesgue) measurable set (of $\mathbb{R}$) easily implies the existence of a bounded measurable set. A proof is as follows - Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = \bigsqcup_{n \in \mathbb{Z}} X \cap (n,n + 1]$. If all $X \cap (n,n+1]$ are measurable, then $X$ must also be measurable, at it is a countable union of measurable sets ($\mathsf{AC}$ is used here). Thus, there exists an $n \in \mathbb{Z}$ in which $X \cap (n,n+1] \subseteq (n,n+2)$ is not measurable.

However, it appears to not be so clear if we only work in $\mathsf{ZF}$, since we cannot guarantee that the countable union above is measurable. I also can't seem to see an easy way around choice here. Thus, if we write:

1. $\mathsf{NM}$ as "there exists a non-measurable set".
2. $\mathsf{BNM}$ as "there exists a bounded non-measurable set".
3. $\mathsf{M}_\omega$ as "countable union of measurable sets is measurable".

My questions are:

> 1. Is $\mathsf{ZF} + \neg\mathsf{M}_\omega$ consistent?
> 2. Is $\mathsf{ZF} + \mathsf{NM} + \neg\mathsf{BNM}$ consistent?