Consider the function over [0,1] given by $f(x)= \begin{cases} 0 & x \in \mathbb{Q}\\ x & x \notin \mathbb{Q} \end{cases}$
In order to compute the Lebesgue integral of $f$ we need to find an increasing sequence of simple functions which are all less than or equal to and converge to $f$.
I was thinking if the function $f_{n}(x)= \frac{1}{n}\lfloor nx \rfloor$ , $x \notin \mathbb{Q}$, would work.
Thanks in advance!
On
Let $\{r_n\}$ be an enumeration of $\mathbb Q$ and define $$\varphi_n = \sum_{k=1}^n r_k\chi_{\{r_k\}}.$$ Then $\varphi_n\leqslant\varphi_{n+1}$ and $\lim_{n\to\infty}\varphi_n=f$, so for each $n$, \begin{align} \int\varphi_n\ \mathsf d\mu &= \int\sum_{k=1}^n r_k\chi_{\{r_k\}}\ \mathsf d\mu\\ &= \sum_{k=1}^n \int_{\{r_k\}} r_k\ \mathsf d\mu\\ &= \sum_{k=1}^n r_k\mu\left(\{r_k\}\right)\\ &= 0. \end{align} It follows that $$\int f\ \mathsf d\mu = \lim_{n\to\infty}\int \varphi_n\ \mathsf d\mu = 0. $$
Did you mean the function $f_n(x)=\frac1n \lfloor nx\rfloor$ if $x\notin \Bbb Q$ and $0$ otherwise? It's the step function with $n$ steps with all rationals set to $0$.
If so, yes, that works. Each step covers a set of the form $[\frac in,\frac{i+1}n]\backslash\Bbb Q$, which is Lebesgue measurable.