I‘m not very versed in Gödel‘s incompleteness theorem but in a naive way: If a statement of existence is not provable, there you cannot find an example which fulfills the statement (otherwise the statement would be provable with this example). But when there is no element which fulfills the statement, doesn’t that imply that the statement ist false?
I thought about that one in the context of the measure problem - because the statement
$$\exists \text{ measure function } \mu: 2^{\mathbb R} \to [0,\infty] \, \forall I = [a,b] \subseteq \bar{\mathbb R}: \mu(I) = b - a$$
is neither provable nor refutable. But if I cannot prove there is a measure function, I cannot find a $\mu$ for which the statement is true. Because finding such a $\mu$ would prove the statement. But when there is no such $\mu$, the statement of existence is false, isn‘t it? Where is my mistake in thinking?
It's not correct that if a statement of existence isn't provable, then you can't find an example that fulfills the statement. It means the theory has at least one model in which you can't find an example to fulfill the statement. But for the statement to be false, you need to be able to say that you can't find an example to fulfill the statement in any model of the theory.
To take a simple example, in the theory of fields with characteristic $0$, the statement $\exists x~(x^2+1 =0)$ is neither provable nor refutable. That just means in some models (fields) it's false (e.g., $\Bbb R$) but in others (e.g., $\Bbb C$), it's true.
If a sentence $\sigma$ is neither provable nor refutable from a theory $T$, that means the theory has at least one model in which the sentence is true and it also has at least one model in which it's false. That's because both $T \cup \{ \sigma \}$ and $T \cup \{ \lnot \sigma \}$ are consistent so by the Completeness Theorem, they both must have models.