Let $\kappa$ be an uncountable cardinal and $(P(\kappa),\cap,\cup, ^c,\kappa,\emptyset)$ the Boolean Algebra of all subsets of $\kappa$. 


**Fact:** If there exists a countably complete *non-principal* ultrafilter on $P(\kappa)$, then $\kappa$ is larger than or equal to a measurable. 


**My question**: Is there a Boolean Algebra $\mathcal{A}$ of subsets of $\kappa$, $\mathcal{A}$ of size strictly less than $2^\kappa$, and the existence of a countably complete *non-principal* ultrafilter on $\mathcal{A}$ would imply that $\kappa$ is (added: larger than or equal to) a measurable/inaccessible/strong limit?


I am interested even in special cases, e.g. $\kappa$ is regular. 



On measurable cardinals
Your question is very interesting in the case $\kappa^+<2^\kappa$. 


For this case, recall that a cardinal $\kappa$ is [weakly measurable](http://cantorsattic.info/Weakly_measurable) if every family of $\kappa^+$ many subsets of $\kappa$ admits a $\kappa$-complete nonprincipal filter measuring them. This notion was introduced and studied by my student Jason Schanker in his dissertation. There are a variety of equivalent characterizations. 


Jason proved that if $2^\kappa$ is larger than $\kappa^+$, then this concept is not necessarily equivalent to measurability, for he constructed models of ZFC with a weakly measurable cardinal, but no measurable cardinal. 


In such a model, therefore, where $2^\kappa=\kappa^{++}$, we have measures for all the families of sets of size smaller than $2^\kappa$, but $\kappa$ is not measurable. So this provides a negative answer to your question. 


Meanwhile, weakly measurable cardinals are necessarily weakly compact, inaccessible and much more, so there is a positive answer to the version of your question at the end where you ask for less than measurability. 


Indeed, the existence of $\kappa$-complete nonprincipal filters measuring families of $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact. In this sense, weak measurability is a generalization of weak compactness from $\kappa$ to $\kappa^+$. 


Thus, if $2^\kappa=\kappa^+$, the answer to your question is no, since the existence of such filters for families of at most $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact, which is a strictly weaker large cardinal concept than measurability. 


