TITLE: Proving ZFC results using large cardinals
QUESTION [9 upvotes]: There are many $ZFC$ results that their proof uses forcing: The idea is that we force the statement to be true, and then using absoluteness (or other reasons) we conclude that the result is true in $ZFC$.
Question. Are there any $ZFC$ results such that in their proof we use large cardinals (and then remove the use of that large cardinal by some arguments like absoluteness)?

REPLY [4 votes]: I am not sure if this is a sort of example you asked for but it seems interesting so I will mention it. There are several ZFC results that are proved in the following way: You assume the result does not hold and that gives you a highly saturated ideal (over a small large cardinal which Fremlin calls quasi measurable). Then you force with this ideal, form the well founded generic ultrapower in the extension and argue towards a contradiction. An example of such a theorem is: For every sequence $\{X_n : n < \omega\}$ of subsets of $[0, 1]$ there is a disjoint refinement of full outer measure, i.e., there is a sequence $\{Y_n : n < \omega\}$ of pairwise disjoint sets such that $Y_n \subseteq X_n$ and $X_n$ and $Y_n$ have same (Lebesgue) outer measure. This is due to Gitik and Shelah. Another such application is: For every $X \subseteq [0, 1]$, there is a $Y \subseteq X$, such that $X$ and $Y$ have same outer measure and the distance between any two points of $Y$ is irrational - So $Y$ is Vitali like inside $X$ and it has full outer measure.