To start, this isn't really Thomae's function, because the interesting bit occurs when the argument is rational. But I couldn't think of any more closely-related function, and Thomae's was close enough.
Anyway, my question:
Let $- \infty < a,b < \infty$ be real numbers and say: $$ T(x) = \begin{cases} t(x) & \text{if $x$ is irrational.} \\ \infty & \text{if $x$ is rational.} \end{cases} $$ Where $t(x)$ is a continuous real-valued function. Then, is $\int_a^bT(x)dx$ defined?
My thoughts: I think this question depends on whether or not we can say $T(x)$ 'equals' infinity at rational $x$ in which case we have some $\infty \times0$ wibble-wobble going on, or we must instead say $T(x)=k$ at rational $x$ and the integral in question becomes $\lim_{k \to \infty}\int_a^bT(x)dx$ which, because the rationals are of density $0$ in $\mathbb{R}$, reduces to $\int_a^bt(x)dx$, which is certainly defined.