Understanding Lemma 2.1 in P. Erdos and I. Z. Ruzsa's "On the Small Sieve. I, Sifting by Primes. I am reading a paper of P. Erdos and I. Z. Ruzsa and I had a question on Lemma 2.1.
Let $A$ be a set of natural numbers.
Let $B$ denote the set of natural numbers divisible by no element of $A$.
Here is Lemma 2.1:

For all $y$ we have:
$$\sum_{b \le y \text{ and } b \in B}1/b \ge \prod_{a \in A}(1 - 1/a)\log(y+1)$$

I am lost at how this relates to the argument which consists of the following points:

*

*Every number has (one or more) decompositions of the form:
$$a_l^{a_l}\dots a_k^{a_k}b,\,\,\,\,b\in B,\,\,\,\,a_i\in A$$


*Hence:
$$\sum_{n \le y}1/n \le \sum_{b \le y \text{ and } b \in B}1/b\prod_{a\in A}(1 + a^{-1} + a^{-2}+\dots)$$


*which immediately yields Lemma 2.1


*Note:  As a by-product, this gives us a proof for the Heilbron-Rohrbach Inequality:
$$\Delta(A) = \lim_{x \to \infty}\frac{F(x,A)}{x} \ge \prod_{a \in A}(1 - 1/a)$$
where $F(x,A)$ denotes the number of natural numbers $n \le x$ divisible by no element of $A$.
I am not clear how the argument leads to the proof of the lemma.  I am unclear how the argument leads to the Heilbronn-Rohrbach Inequality.
Any explanation or hint would be highly appreciated.
 A: The inequality 
$$
\sum_{n \le y}1/n \le \sum_{b \le y \text{ and } b \in B}1/b\prod_{a\in A}(1 + a^{-1} + a^{-2}+\dots)
$$
follows from admitting the expression for $n$ as $a_l^{a_l}\dots a_k^{e_k}b,\,\,\,\,b\in B,\,\,\,\,a_i\in A$ for any possible $a_i \in A$, $e_i$'s and $b\in B$. Thus, possibly over-counting happens, but the inequality is true. One way to understand the right side is to convert the sum over those expression into a product.
A few things that you need to know are
$$
\log(y+1)\leq \sum_{n\leq y} \frac1n
$$
This can be shown by comparing the graphs of $f(x) = \frac1{x+1}$ and $g(x) = \frac1{\lfloor x+1 \rfloor}$. Since we have $f(x)\leq g(x)$ for $0\leq x \leq y$, we obtain the result by integration:
$$
\log(y+1)=\int_0^y \frac1{x+1}dx \leq \int_0^y \frac1{\lfloor x+1 \rfloor} dx = \sum_{n\leq y} \frac1n 
$$
with the last equality is true for any positive integer $y$. 
Secondly, by the geometric series
$$
\frac1{1-r}=1+r+r^2+\cdots , \ \ \ |r|<1,
$$
we have
$$
\prod_{a\in A} (1+a^{-1}+a^{-2}+\cdots) = \prod_{a\in A} \frac1{1-\frac1a}=\left(\prod_{a\in A}\left(1-\frac1a\right)\right)^{-1}.
$$
Then we have
$$
\log(y+1)\leq \sum_{b\leq y \ \mathrm{and} \ b\in B} 1/b \prod_{a\in A} \left(1-\frac1a\right)^{-1}.
$$
This yields the lemma:
$$
\sum_{b\leq y \ \mathrm{and} \ b\in B} 1/b \geq \prod_{a\in A}\left(1-\frac1a\right) \log(y+1).$$
For Heilbron-Rohrbach, assume the existence of $\Delta(A)$. Then for any $a\in A$, $e\geq 1$, we have
$$
\lim_{x\rightarrow\infty}\frac{F(x/a^e, A)}{x/a^e} = \Delta(A),
$$
since $x/a^e\rightarrow\infty$ as $x\rightarrow\infty$. Then we set up the counting as below: with $a_i\in A$, $e_i\geq 1$, $b\in B$. For any $\epsilon>0$ and sufficiently large $x$, we have
$$
\sum_{n\leq x} 1 \leq \sum_{\prod a_i^{e_i} b \leq x} 1=\sum_{a_i, e_i} F\left(\frac x{\prod a_i^{e_i}},A\right)\leq (\Delta(A)+\epsilon)\sum_{a_i,e_i} \frac x{\prod a_i^{e_i}}
$$
The last sum is $x\prod_{a\in A}(1+a^{-1}+a^{-2}+\cdots)$. As before, this is $x\prod_{a\in A} (1-1/a)^{-1}$. Then the result follows by taking $\epsilon\rightarrow 0+$. 
A: For proving this you really need :


*

*$1+a^{-1}+a^{-2}+\ldots = \frac{1}{1-1/a} $.

*$$\prod_{a \in A} (1+a^{-1}+a^{-2}+\ldots) = \sum_{c \in E^A} c^{-1}$$ where $E^A$ is the set whose elements are of the form $a_l^{e_l}\dots a_k^{e_k}$. If $A$ contains $2,4,8$ then $4$ has $4$ different factorization, in that case repeat it $4$ times in $E^A$.

*$\sum_{n=1}^N n^{-1} \ge \int_1^N \frac{dx}{x} = \log(N)$
