(Edited for more specific question)
I'm looking for a generalization of proof of the implication between "there exists a measurable cardinal $\kappa$" and "$0^\sharp$ exists". The proof I know uses Rowbottom's theorem:
Theorem (Rowbottom). If $\kappa$ is measurable, then given $f_i:[\kappa]^{m_i}\to\omega$ for $i<\omega$ and $m_i<\omega$ there is a subset $H\subseteq\kappa$ of cardinality $\kappa$ such that every $f_i$ is constant on $[H]^{m_i}$.
And then I can build a club of indiscernibles for the model $\langle L_{\omega_1}, \in\rangle$.
My question is: how can I generalize the proof to any set $a$, to have an implication from "there exists a measurable cardinal $\kappa>\lambda$" to "$a^\sharp$ exists, for all $a\subseteq\lambda$"? I think we can start by this generalization of Rowbottom's theorem:
Theorem (Rowbottom). If $\kappa$ is measurable and $\lambda<\kappa$, then given $f_\alpha:[\kappa]^{m_\alpha}\to\lambda$ for $\alpha<\lambda$ and $m_\alpha<\omega$ there is a subset $H\subseteq\kappa$ of cardinality $\kappa$ such that every $f_\alpha$ is constant on $[H]^{m_\alpha}$.
But then I don't see how to build a club of indiscernibles for $\langle L_{\omega_1}[a], \in, a\cap L_{\omega_1}[a]\rangle$.
(Previously I asked if there was a simpler proof than passing through Rowbottom's theorem, but I guess it wasn't exactly what I was looking for)