I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
2.39 condition for measurable function
Suppose $(X,\mathcal{S})$ is a measurable space and $f:X\to\mathbb{R}$ is a function such that $$f^{-1}((a,\infty))\in \mathcal{S}$$ for all $a\in\mathbb{R}$. Then $f$ is an $\mathcal{S}$-measurable function.
$\mathcal{T}=\{A\subset\mathbb{R}:f^{-1}(A)\in\mathcal{S}\}.$
In the proof of 2.39, the author wrote as follows:
By hypothesis, $\mathcal{T}$ contains $\{(a,\infty):a\in\mathbb{R}\}$. Because $\mathcal{T}$ is closed under complementation, $\mathcal{T}$ also contains $\{(-\infty,b]:b\in\mathbb{R}\}$. Because the $\sigma$-algebra $\mathcal{T}$ is closed under finite intersections (by 2.25), we see that $\mathcal{T}$ contains $\{(a,b]:a,b\in\mathbb{R}\}$. Because $(a,b)=\bigcup_{k=1}^\infty (a,b-\frac{1}{k}]$ and $(-\infty,b)=\bigcup_{k=1}^\infty (-k,b-\frac{1}{k}]$ and $\mathcal{T}$ is closed countable unions, we can conclude that $\mathcal{T}$ contains every open subset of $\mathbb{R}$.
I wonder why the author wrote the fact $(-\infty,b)\in\mathcal{T}$ in his proof.
If we know $(a,b)\in\mathcal{T}$, then we can conclue that $\mathcal{T}$ contains every open subset of $\mathbb{R}$ because $\{(a,b):a,b\in\mathbb{Q}\}$ is a basis of $\mathbb{R}$ and $\{(a,b):a,b\in\mathbb{Q}\}$ is countable.
The author thought that it was simpler to avoid using the fact that $\{(a,b):a,b \in \mathbb{Q}\}$ is a countable basis for $\mathbb{R}$. Defining the basis and saying it's a countable basis for $\mathbb{R}$ requires the reader to check more facts (check that it's countable, check that it's a basis, check the definition of a basis) than just checking the claim $(-\infty,b)=\bigcup_{k=1}^\infty (-k,b-\frac{1}{k}]$ and that the right-hand side is a countable union.