Context:
I'm reading theorem which requires me to check the following conditions:
Let the vector valued function $f(t,x)$ be defined almost everywhere and measurable in the domain $G$ of the $(t,x)$-space ($t\in[0,\infty), x\in\mathbb{R}^n$) and let there exists, for each bounded closed domain $D\subset G$, an almost everywhere finite function $m(t)$ such that $$ |f(t,x)|\leq m(t) $$ almost everywhere in $D$.
and the theorem continues with some conclusions on $f(t,x)$ and other objects arising from it, which I omit here since those are not important for this question. However, I wan't to use that theorem for the following $f(t,x)$: $$ f(t,x) = \frac{x}{T-t} $$ with $n=1,G=\mathbb{R}, T>0$. So I need to check if my $f(t,x)$ is defined almost everywhere and measurable for any $x\in\mathbb{R}$ and if there exists the function $m(t)$. So, for each closed bounded domain $D\in\mathbb{R}$ with $\inf D=\underline{x}, \sup D = \overline{x}$, I proposed $$ m(t) = \frac{\max(|\underline{x}|,|\overline{x}|)}{T-t} $$ which verifies $|f(t,x)|\leq m(t)$. Now, I have to check that this function is almost everywhere finite (in $t$), which I believe it is, since it is finite except for $t=T$. However, I'm not entirely sure I'm missing something, or misinterpreting what "almost everywhere finite" means.
Question:
So, Is it correct to say that $\frac{1}{T-t}$ is almost everywhere finite in the context of the question?