Approximation Algorithm for the Unique Coverage Problem I was reading about the $\mathcal{O}(\frac{1}{\log n})$ approximation algorithm for the Unique Coverage Problem from these notes.

The gist of the algorithm is as follows:


  
  Partition the elements into $\log n$ classes according to their degrees, i.e., the number of sets that cover the element. So class $i$ contains all the elements which are covered with at least $2^i$ and at most $2^{i+1}$ sets.
  Let $i$ be the class of the maximum cardinality.
  Choose any set with probability $\frac1{2^i}$
  


Then in Lemma 3, they go on to show that


  Lemma 3: The expected number of the elements uniquely covered from class $i$
  is $\frac{1}{e^2}$ fraction of the element in class $i$.


I was able to follow the algorithm till here but I am completely stumped as to how they arrived at the statement immediately following lemma 3


  Therefore, the total profit of uniquely covered elements is at least
  $\frac{1}{e^2 \log n} \times \text{opt}$


How did they arrive to this conclusion? They didn't even show what the upper bound is for this maximization algorithm. What am I overlooking?

It is trivial to prove that the expected number of unique elements covered by the algorithms is $\frac{n}{e^2}$ since it directly follows Lemma 3 but that is about all that I am able to observe :/
Suppose there are $m$ elements and $n$ sets (this is the meaning of your parameter $n$). The algorithm divides the set of elements into $\log n$ classes, so one of these classes, say class $i$, contains at least $m/\log n$ elements. Lemma 3 shows that if you pick each set with probability $1/2^i$, then in expectation you uniquely cover a $1/e^2$ fraction of the elements in class $i$. Thus, you uniquely cover at least $m/(e^2\log n)$ elements in expectation. On the other hand, the optimal cover uniquely covers at most $m$ elements. Thus the approximation ratio of this algorithm is at least $1/(e^2\log n)$.