Job scheduling approximation
The instance in the example is "almost" the right one, however you are right that, as given, it does not prove tightness. The instance is missing two more jobs of size $m$. 

After we include these two jobs, we have 2 jobs of each size from $2m-1$ to $m+1$ and 3 jobs of size $m$ (for $n=2m+1$ jobs in total, as stated). 

Under the modified greedy algorithm, the maximum load will be $4m-1$. However, one can instead schedule a job of size $2m-i$ with a job of size $m+i$ for each $i=1,2,..., \lfloor  m/2 \rfloor$, and the remaining 3 jobs of size $m$ separately on the same machine. The maximum load will then be $3m$.