Let $k$ be a number field and let $k_\infty$ be the **cyclotomic** $\mathbf{Z}_p$-extension of $k$. Put $\Gamma=G(k_\infty/k)\cong \mathbf{Z}_p$, $\Lambda=\mathbf{Z}_p[[\Gamma]]$. Let $S$ be a finite set of primes of $k$ which contains primes above p and infinite primes. For simplicity, assume $p\neq 2$. Put

$L_\infty=$ the maximal unramified abelian pro-$p$-extension of $k_\infty$, $\quad$ $X=G(L_\infty/k_\infty)$;

$L_\infty^S=$ the maximal $S$-ramified abelian pro-$p$-extension of $k_\infty$, $\quad$ $X_S=G(L_\infty^S/k_\infty)$.

As usual, both $X$ and $X_S$ are finitely generated $\Lambda$-module ($X$ is even torsion). Thus one can talk about their $\mu$-invariants. My question is, is it true that $\mu(X)=\mu(X_S)$? (Or weaker: Is it true that $\mu(X)=0\Longleftrightarrow \mu(X_S)=0$?)

I know that when $k$ contains $\mu_p$ ($p$-th roots of unity), then $\mu(X)=\mu(X_S)$. (Ref. Cohomology of Number Fields, XI. $\S 3$). But what if $\mu_p\not\subseteq k$?

Another question. Suppose that $K/k$ be a finite extension, unramified outside $S$. Then it is easy to prove $\mu(X(k))\le \mu(X(K))$. (Ref. Washington's book (2nd Ed.), $\S 7.5$) Is it true that $\mu(X_S(k))\le \mu(X_S(K))$?