TITLE: Does asserting the existence of large cardinals allow one to prove the existence of new subsets of $\mathbb{R}$?
QUESTION [7 upvotes]: ZFC proves the existence of certain subsets of $\mathbb{R}$. Suppose ZFC is consistent, and we adjoin a large cardinal axiom to ZFC, obtaining ZFC'. Assume ZFC' is also consistent. It it possible to prove the existence of new subsets of $\mathbb{R}$ using ZFC'?

REPLY [3 votes]: It would be more precise to say that "ZFC + large cardinals" proves more statements positing the existence of sets of reals with certain logical properties (e.g. it proves more $\Sigma^2_1$ statements) than to say that it proves the existence of new subsets of $\mathbb{R}$. That being said, assuming the existence of large cardinals certainly does allow you to prove more along these lines, even about single reals.
For example, if there is a measurable cardinal then $0^\sharp$ exists, i.e. there is a unique real $x$ that computes the set of sentences true in Goedel's constructible universe $L$ (and a bit more.)  The statement $x = 0^\sharp$ is absolute between models of set theory containing a given real $x$ and having the same ordinals, so $0^\sharp$ is a fairly robust notion.
However, $0^\sharp$ is not contained in the inner model $L$ (which satisfies ZFC) so one cannot prove the statement "$0^\sharp$ exists" in ZFC alone.
If you want to consider sets of reals rather than single reals, then you could consider the statement "$\mathbb{R}^\sharp$ exists", which is stronger but still follows from the existence of a measurable cardinal.  Some important but technical $\Sigma^2_1$ statements that are even stronger than this in terms of consistency strength are iterability statements like "there exists an iteration strategy for a premouse with a superstrong cardinal."