Why does the definition of limit imply these two facts in the result on the limit of measurable functions in Measure, Integration & Real Analysis?

I have two potentially related questions concerning limits. I'm reading 2.48 in Measure, Integration & Real Analysis by Sheldon Axler, which is reproduced below.

1. Why does the definition of limit imply that there exists $m \in \mathbb{Z}^+$ such that $f_k(x) > a + \frac{1}{j}$ for all $k \geq m$? I know the definition of the limit of a sequence but am having trouble using it to clarify this implication.

2. In the proof of the inclusion in the other direction, why is it $f(x) \geq a + \frac{1}{j}$ and not $f(x) > a + \frac{1}{j}$ when taking the limit as $k \to \infty$?

I know these are basic but I've tried fiddling with these facts for a bit of time and am still unsure of my reasoning. Thanks in advance for considering my question!

2.48 Limit of $\mathcal{S}$-measurable functions. Suppose $(X, \mathcal{S})$ is a measurable space and $f_1, f_2, \dots$ is a sequence of $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$. Suppose $\lim_{k \to \infty} f_k(x)$ exists for each $x \in X$. Define $f: X \to \mathbb{R}$ by$$f(x) = \lim_{k \to \infty} f_k(x).$$Then $f$ is an $\mathcal{S}$-measurable function.

Proof. Suppose $a \in \mathbb{R}$. We will show that$$f^{-1}((a, \infty)) = \bigcup_{j=1}^{\infty}\bigcup_{m=1}^{\infty} \bigcap_{k=m}^{\infty} f_k^{-1}\left(\left(a + \frac{1}{j}, \infty\right)\right),\quad\quad 2.49$$which implies that $f^{-1}((a, \infty)) \in \mathcal{S}$.

To prove 2.49, first suppose $x \in f^{-1}((a, \infty))$. Thus there exists $j \in \mathbb{Z}^+$ such that $f(x) > a + \frac{1}{j}$. The definition of limit now implies that there exists $m \in \mathbb{Z}^+$ such that $f_k(x) > a + \frac{1}{j}$ for all $k \geq m$. Thus $x$ is in the right side of 2.49, proving that the left side of 2.49 is contained in the right side.

To prove the inclusion in the other direction, suppose $x$ is in the right side of 2.49. Thus there exist $j, m \in \mathbb{Z}^+$ such that $f_k(x) > a + \frac{1}{j}$ for all $k \geq m$. Taking the limit as $k \to \infty$, we see that $f(x) \geq a + \frac{1}{j} > a$. Thus $x$ is in the left side of 2.49, completing the proof of 2.49. Thus $f$ is an $\mathcal{S}$-measurable function.