Let $E$ be countably infinite and $v$ a probability measure on it with support $C\subseteq E$, i.e. $v(E)=v(C)=1$.
Let $u$ be another measure on E with $u(E)\leq 1$.
Now it was shown that $v(i)\leq u(i)$ for all $i\in C$.
From this it follows that $v(E)\leq u(E)\implies u(E)=1$.
Then it is said that in particular this says that the measure $u$ on $E$ is unique. But why?
The measure $u$ is unique because it must be equal to $v$. We are given $v(A) \leq u(A)$. Now let us see that $u(A) \leq v(A)$.
We have $u(A) = 1- u(A^c) \leq 1- v(A^c) = v(A)$ where in the last equality we used that $v(E)=1$.
Thus $u(A) = v(A)$.