We know that if $\kappa$ is a measurable cardinal and $\mu$ be a two-valued non-trivial$\kappa$-additive measure on it then the corresponding inner model produced by Mostowski collapse of Scott's ultraproduct ($M_{\kappa,\mu}$) is dependent on both $\kappa$ and $\mu$. So by changing measures on a certain measurable cardinal we can produce many different inner models. Even by transitivity if two such inner models be isomorphic then they are equal. Now a question is:

"Is it possible to have a measurable cardinal which its inner models be independent from its measures?"

Precisely:

**Definition (1):** A "unique measurable cardinal" is a measurable cardinal which for all two-valued non-trivial $\kappa$-additive measures like $\mu$ and $\mu'$ on it we have $M_{\kappa , \mu}=M_{\kappa , \mu'}$. We show this unique inner model by $M_{\kappa}$.

**Question (1):** Is the following statement true?

$Con(ZFC+ \text{there exists a measurable cardinal})\Longrightarrow$ $Con(ZFC+ \text{there exists a unique measurable cardianl})$

Now the main question is about the behavior of the "well defined" function $\kappa \mapsto M_{\kappa}$:

**Definition (2):** Define:

The collection of all unique measurable cardinals:

$UM:=\lbrace \kappa~|~\kappa~\text{is a unique measurable cardinal}\rbrace$

The (informal) collection of all inner models of unique measurable cardinals:

$IUM:=\lbrace M_{\kappa}~|~\kappa \in UM \rbrace$

**Question (2):** Is the function $\kappa \mapsto M_{\kappa}$ from $UM$ to $IUM$ (strictly) increasing? In the other words which one of the following statements are true?

$(1)~\forall \kappa,\lambda \in UM~~~~~(\kappa < \lambda \longrightarrow M_{\kappa}\subseteq M_{\lambda})$

$(2)~\forall \kappa,\lambda \in UM~~~~~(\kappa < \lambda \longrightarrow M_{\kappa}\subsetneq M_{\lambda})$

I thinkis that $M_\kappa$ and $M_\lambda$ are not subsets of each other at all. In $M_\kappa$, $\lambda$ is still measurable and in $M_\lambda$, $\kappa$ is measurable. I'm not 100% behind this argument, though. $\endgroup$