The relation between uniform integrability and dominated convergence theorem in the case of counting measure

If you strictly follow the definition, any integrable family of functions on $\Bbb{Z}_{\geq 0}$ should be uniformly integrable with respect to counting measure, since $\delta = 1$ works for any $\epsilon > 0$, as a set of measure less than $1$ must have measure $0$.

"Is $(\mathbb{N},P(\mathbb{N}),\mu)$ a finite measure space?" It is not. I think this is basically the problem , since with the counting measure you could take $\delta = 1/2$ for all $\epsilon$, and then the condition is vacuous true, while if you take any finite measure subset convergence issues go away.

If you want to work with non-finite measure spaces, you need this theorem: https://en.wikipedia.org/wiki/Vitali_convergence_theorem

A good example to think about is the sequence of sequences $(1_{n})_{n \geq 1}$ ( a point mass moving towards infinity).

Another good example is to consider the sequence of sequences, where the $nth$ sequence is $\alpha_n = \frac{1}{n} ( 1,\frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{2^{n^2}}, 0 ,0 ,\ldots)$. What does Vitali's convergence theorem say about this?