The relation between uniform integrability and dominated convergence theorem in the case of counting measure I want to discover the analog between uniform integrability (UI) and the dominated convergence theorem (DCT) for infinite series.
An infinite series is an integral with respect to counting measure. That is, $\sum_{k=0}^{\infty}f(k)=\int_{\mathbb{Z}_{\geq 0}}f(k)\mu(dk),$ where $\mu$ is the counting measure.  We can then say a function $f$ is integrable with respect to the counting measure if $\sum_{k=0}^{\infty}|f(k)|=\int_{\mathbb{Z}_{\geq 0}}|f(k)|\mu(dk)<\infty.$
My first step is to reformulate DCT and UI.
For DCT, we are consider a sequence of $\mu-$integrable function $f_{n}(k)$ such that $f_{n}(k)\longrightarrow f(k)$ as $n\rightarrow\infty$, and for each $n$, $|f_{n}(k)|\leq g(k)$ for some $\mu-$integrable $g(k)$. Then, $$\lim_{n\rightarrow\infty}\sum_{k=0}^{\infty}f_{n}(k)=\sum_{k=0}^{\infty}\lim_{n\rightarrow\infty}f_{n}(k)=\sum_{k=0}^{\infty}f(k).$$
However, I don't know how to formulate the UI. The definition UI I have is that:

A family $\{f_{\alpha\}}$ of integrable function is uniform integrable if for any $\epsilon>0$, there is a $\delta>0$ such that whenever $\lambda(A)<\delta$, $\sup_{\alpha}\int_{A}|f_{\alpha}|\lambda(dx)<\epsilon,$ where $\lambda$ here is a general measure.

I can reformulate this without problem, but I don't understand what does it mean for $\mu(A)<\delta$ when $\mu$ is a counting measure.
I read some online notes saying that DCT is a consequence of UI and Egoroff Theorem, I understand this, since Egoroff will give you a set $A$ on which the convergence is uniform and UI can make the integral on $A^{c}$ to be negligible. But this requires the measure space to be finite. I am not sure if we can apply this to counting measure, i.e. is $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$ a finite measure space?
Even I can answer this question, I don't want to stop here.I believe that  there must be something special in the case of counting measure, since $\mu(A)<\delta$ in the case of counting measure is still mysterious to me, but for now I don't know where to continue.
Thank you!

Edit: Example
Okay, I worked some examples, but I still don't quite understand what fails.
For example, consider the sequence $f_{n}(x)$ on the integer $X:=\{1,2,3,\cdots\}$, defined by
$$f_{n}(x)=\left\{
  \begin{array}{ll}
\frac{1}{n},\ \ \text{if}\ \ x=1,2,\cdots, n\\
0,\ \ \text{if}\ \ x\geq n+1.
\end{array}
\right.$$
Note that $\sum_{k=1}^{\infty}f_{n}(x)=1$ for every $n$, but $\lim_{n\rightarrow\infty}f_{n}(x)=0$ for every $x$, so we cannot interchange the limit and summation.
As suggested by the comment, uniform integrability is guaranteed in the case of counting measure, so What fails here?

Edit: Potential Answer
As Rivers Mcforge said, the example above does not satisfy the boundedness requirement in DCT. Along with the comment of Lorenzo, I found a connection between the boundedness requirement and tightness.
As Lorenzo suggested, the sequence in above example is not $\mu-$tight, and since we are over an infinite measure space, the Vitali convergence theorem needs the tightness. (The uniform integrability was given to us for free over our space, as suggested by both answers below).
Eventually, I found that in the case of our counting measure space, a sequence that satisfies DCT will also satisfy Vitali. In other words, we can use Vitali to prove DCT (in the case of our counting measure space $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mu)$).
Indeed, recall that for any measure $\lambda$ and any measurable set $E$ (not necessarily of finite measure), if $f$ is $\lambda-$integrable over $E$, then for each $\epsilon>0$, there is a set of finite measure $E_{0}$ for which $$\int_{E\setminus E_{0}}|f(x)|\lambda(dx)<\epsilon.$$
Now, suppose $f_{n}(x)$ is an sequence of functions on $X$ that converges $\mu-$almost surely to $f(x)$. It is for free that $f_{n}(x)$ is uniformly integrable. If $|f_{n}(x)|\leq g(x)$ for all $n$ and $x$, where $g(x)$ is $\mu-$integrable, then by the above recalled fact, for each $\epsilon>0$, there is a subset $X_{0}$ of $X$ of finite measure for which $$\int_{E\setminus E_{0}}|f_{n}(x)|\mu(dx)\leq \int_{E\setminus E_{0}}|g(x)|\mu(dx)<\epsilon,\ \text{for all}\ n.$$ Hence, the family $\{f_{n}(x)\}$ is tight.
Then, it follows from Vitali that we can interchange the summation (the $\mu-$integral) with the limit of $n\rightarrow\infty.$ Thus, the confusion of DCT follows.
I am not sure if tightness can conversely imply the boundedness requirement in DCT.
 A: 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 $0$.
A: "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?
