Number of squarefree integers $n \le x$ with $\omega(n)$ even Let $X$ be the set of positive squarefree integers. It is well known that $X$ admits asymptotic density $6/\pi^2$, i.e. 
$$
|X\cap [1,x]|=\frac{6}{\pi^2}x(1+o(1)),\,\,\,\,   \text{ as }x\to \infty.
$$
Let $S$ be the subset of $X$ of integers with an even number of distinct prime factors. Is it true that $S$ admits asymptotic density too?
Ps. The question is related to the Mertens' function $M(x)=\sum_{n\le x}\mu(n)$. In particular $S$ admits asymptotic density if and only if also $S^c$ has it. In that case, it would imply $M(x)=cx(1+o(1))$ for some real constant $c$. Here, we just recall that the RH is equivalent to $M(x)=O(x^{1/2+\varepsilon})$ for all $\varepsilon>0$. Moreover, the PNT is equivalent to $M(x)=o(x)$. Is it enough to conclude?
 A: As a brief comment  observe that for $(-1)^{\omega(n)}$ for squarefree
integers we have the Euler product
$$L_1(s) = \prod_p \left(1-\frac{1}{p^s}\right)
= \sum_{n\ge 0} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}.$$
Furthermore we have
$$L_2(s) = \prod_p \left(1+\frac{1}{p^s}\right)
= \prod_p \frac{1-1/p^{2s}}{1-1/p^s}
= \frac{\zeta(s)}{\zeta(2s)}.$$
Therefore the Dirichlet series of the indicator of squarefree integers
having an even number of prime factors is given by
$$\frac{1}{2} (L_1(s)+L_2(s))
= \frac{1}{2} \left(\frac{1}{\zeta(s)} + 
\frac{\zeta(s)}{\zeta(2s)}\right).$$
Now               apply               the              Wiener-Ikehara
theorem
to get for the density
$$\frac{x^1}{1} \mathrm{Res}_{s=1}
\frac{1}{2} \left(\frac{1}{\zeta(s)} + 
\frac{\zeta(s)}{\zeta(2s)}\right)
= \frac{3}{\pi^2} x.$$
We may or may  not be allowed to bring the full  power of this theorem
to the question.
A: This can be done with Selberg-Delange method. This method is well-explained in Montgomery and Vaughan, 'Multiplicative Number Theory', I. Classical Theory, also in Tenenbaum's 'Introduction to Analytic and Probabilistic Number Theory'. 
By Theorem 7.18 in Montgomery and Vaughan, we obtain the following:
Let $R>2$. Let $F(s,z)$ be defined by
$$
F(s,z)=\prod_p \left(1+\frac z{p^s}\right) \left(1-\frac 1{p^s}\right)^z
$$
for $\mathrm{Re}(s)>\frac12$ and $|z|\leq R$. 
Denote by $a_z(n)=\mu^2(n) z^{\omega(n)}$ the coefficients in the Dirichlet series 
$$
\zeta(s)^z F(s,z) = \sum_{n=1}^{\infty} \frac{a_z(n)}{n^s}, \ \ \sigma>1.
$$
Let $A_z(x)=\sum_{n\leq x} a_z(n)=\sum_{n\leq x} \mu^2(n)z^{\omega(n)}$. We have
$$
A_z(x) = \frac{F(1,z)}{\Gamma(z)}x (\log x)^{z-1} + O\left( x(\log x)^{\mathrm{Re}(z)-2}\right).
$$
We use this for $z=\pm 1$. For $z=1$, it is the well-known result in your question. For $z=-1$, we have
$$
A_{-1}(x)=\frac{F(1,-1)}{\Gamma(-1)}x (\log x)^{-2} + O\left( x(\log x)^{-3}\right).
$$
Although $\Gamma(-1)$ is undefined, we can regard its reciprocal as $0$. ($\Gamma$ function has simple pole at $-1$) 
Thus, we have
$$
A_{-1}(x) = O\left( x(\log x)^{-3}\right).
$$
Therefore, we obtain 
$$
\sum_{n\leq x} \mu^2(n) ( 1 + (-1)^{\omega(n)} ) = \frac6{\pi^2} x + O\left( \sqrt x\right) + O\left( x(\log x)^{-3}\right).
$$
This gives 
$$
\sum_{n\leq x, \ \omega(n) \ \mathrm{ is } \ \mathrm{even}} \mu^2(n) = \frac3{\pi^2} x +  O\left( x(\log x)^{-3}\right). 
$$
Hence, the density of integers $n\leq x$ such that $\omega(n)$ is even, among the square-free integers $n\leq x$ is $\frac12$. 
This method also allows us to have estimates of the type: For $q\geq 3$, 
$$
\sum_{n\leq x, \ \omega(n) \ \equiv a \ \mathrm{mod} \ q} \mu^2(n) = \frac 6{\pi^2} \frac xq + O(x (\log x)^{\cos(\frac{2\pi}q)-1}).
$$
Then Mertens function $M(x)$ according to this asymptotic, becomes 
$$
M(x) =  O\left( x(\log x)^{-3}\right).
$$
On the other hand, this is a stronger form of PNT. Recall that the PNT implies $M(x)=o(x)$. In fact, by Perron's formula, we obtain a stronger estimate
$$
M(x) = O\left( x \exp(-c \sqrt{\log x})\right).
$$
Applying current knowledge on the zero-free region of the Riemann zeta function, we have 
$$
M(x) = O\left( x\exp(-c (\log x)^{\frac35} (\log\log x)^{-\frac15})\right).
$$
For further informations, you may find this interesting.
