Problem: Let the function $f$ be defined such that $$ f(x)=\begin{cases} 1 \quad x\in\mathbb{R}\backslash\mathbb{Q}\\ 2 \quad x\in \mathbb{Q}\end{cases}.$$
Compute the folowing integral: $\quad$$\int_0^1 f(x)\,dx$.
Would it be $1$ because there are infinitely many irrational numbers between each rational number?
I don't even know where to start with this.
On
It could depend on exactly how you are defining integration. Have you heard the phrase "almost everywhere"? It means that the exceptions are a set of measure zero. In the reals, any finite or countable set has measure zero. So, your function is equal to 1 almost everywhere. Even some uncountable set have measure zero; not all, of course.
Riemann integral: $\int_0^1 f(x) dx$ is undefined, because the refinement of the partition of the interval does not result in a converging Riemann sum.
Lebesgue integral: $f\equiv 1$ almost everywhere, because the rational numbers are a countable set, which means that they are a set of measure zero and do not have any impact on the result. Therefore, $\int_0^1 f(x) dx = 1$