Proportionality constant in Montgomery-Vaughan Theorem 7.20

Question

In Multiplicative Number Theory - Vol. I by Montgomery and Vaughan the following result is proved.

Theorem 7.20 Let $A(x,r)$ denote the number of $n\leq x$ such that $\Omega(n)\leq r \log \log x,$ and let $B(x,r)$ denote the number of $n\leq x$ for which $\Omega(n)\geq r \log \log x.$ If $0<r\leq 1$ and $x\geq 2$ then $$ A(x,r)\ll x(\log x)^{r-1-r \log r}. $$ If $1\leq r \leq R<2$ and $x\geq 2$ then $$ B(x,r)\ll_{R} x(\log x)^{r-1-r \log r}. $$ How does the $R$-dependent proportionality constant vary as $R$ ranges over $[1,2)$?

Also, the indicated proof techniques are the same for both $A(x,r)$ and $B(x,r),$ but $r<R$ is required for $B(x,r)$. What happens if $r\geq 2$? Surely then even fewer $n$ satisfy the inequality $\Omega(n)\geq r \log \log n$ so the bound on $B(x,r)$ still holds. So why the restriction $r<R$?

Answer 1

Edit: My only goal is to mark this as answered.

As explained by Greg Martin in a comment:

Certainly $𝐵(𝑥,𝑟)$ is a decreasing function of $𝑟$, but that doesn't imply that the bound on $𝐵(𝑥,𝑟)$ continues to hold for $𝑟\geq 2$.

The reason that $𝑅<2$ is required in the given proof is that it proceeds via upper bounds for $$ \sum_{n \leq x} \Omega(n), $$ and this sum genuinely changes character when $𝑟>2$, since the integers $𝑛$ that are powers of 2 (or nearly so) are then increased by the map $n \mapsto \Omega(n)$ ; for example, the largest power of $2$ less than $𝑥$ already gives a contribution of

$$x^{(\log r)/\log 2}$$ to the sum.