[example: Rough Integers Up To x]
Let $A=\{1,2,\dots,x\}$ with $z\le x$. By the definition of $S(A,P,z)$, an integer $n\le x$ is counted exactly when $\gcd(n,P(z))=1$, which means that no prime $p<z$ divides $n$; equivalently, every prime factor of $n$ is at least $z$.
For a fixed prime $p<z$, the multiples of $p$ in $\{1,\dots,x\}$ are
\begin{align*}
p,2p,3p,\dots,\left\lfloor \frac{x}{p}\right\rfloor p,
\end{align*}
so their number is $\lfloor x/p\rfloor$ and their proportion is $\lfloor x/p\rfloor/x$. Since
\begin{align*}
\frac{x}{p}-1 < \left\lfloor \frac{x}{p}\right\rfloor \le \frac{x}{p},
\end{align*}
we get
\begin{align*}
\frac{1}{p}-\frac{1}{x} < \frac{1}{x}\left\lfloor \frac{x}{p}\right\rfloor \le \frac{1}{p}.
\end{align*}
Thus the proportion divisible by $p$ is $1/p+O(1/x)$, and the proportion not divisible by $p$ is $1-1/p+O(1/x)$. The independence heuristic discards the small endpoint errors and multiplies these local survival proportions over the primes $p<z$, giving
\begin{align*}
S(A,P,z) \approx x\prod_{p<z}\left(1-\frac{1}{p}\right).
\end{align*}
By *Mertens' theorem*,
\begin{align*}
\prod_{p<z}\left(1-\frac{1}{p}\right) \sim \frac{e^{-\gamma}}{\log z},
\end{align*}
so the predicted main scale is
\begin{align*}
x\prod_{p<z}\left(1-\frac{1}{p}\right) \sim \frac{e^{-\gamma}x}{\log z}.
\end{align*}
This is the basic local-product prediction: sieving by primes below $z$ should leave about a constant multiple of $x/\log z$ integers.
[/example]