Alternative representation of the Dirichlet Function?

Question

In general, if two functions $f,g:[a,b]\rightarrow \mathbb{R}$ are Riemann integrable then $h(x)=\max\{f(x),g(x)\}$ is Riemann integrable as well, but this result does not extend to a countable family of Riemann integrable functions. To prove this result, this exercise asks specifically to find a family $\{f_n\}_{n\in \mathbb{N}}$ of step functions defined on $[a,b]$ such that for every $n\in\mathbb{N}$ for every $x\in[a,b]$ the value of $f_n(x)$ is either $1$ or $0$, but with the property that $g(x)=\max\{f_n(x)\}_{n\in \mathbb{N}}$ is not Riemann integrable.
My idea is to show that there is a family $\{f_n\}_{n\in \mathbb{N}}$ such that $$g(x)=\max\{f_n(x)\}_{n\in \mathbb{N}}=\begin{cases} 1, & \text{if }x\in\mathbb{Q}\cap[a,b] \\ 0, & \text{otherwise } \end{cases}$$ which, as is well known, is not a Riemann integrable function. Let $\{r_n\}_{n\ge1}$ be an enumeration of the rationals with $a\leq r_n\leq b$. Define $$f_n(x)=\begin{cases}1, & \text{if }x=r_n \\ 0, & \text{otherwise} \end{cases}$$
Clearly this is a family of step functions since each function is constant except at a single point. Now if $x\in\mathbb{Q}\cap[a,b]\text{ } $ it is clear that $f_{n_0}(x)=1$ for some $n_0 \in \mathbb{N }$ and $f_n(x)=0$ for every $n\neq n_0$. This means that $g(x)=\max\{f_n\}_{n\in\mathbb{N}}=1$. Moreover, if $x\notin \mathbb{Q}\cap[a,b]$ then $f_n(x)=0$ for every $n\in\mathbb{N}$, thus $g(x)=\max\{f_n\}_{n\in\mathbb{N}}=0$. In conclusion$$g(x)=\begin{cases} \max\{f_n(x)\}_{n\in\mathbb{N}}=1, & \text{if }x\in\mathbb{Q}\cap[a,b] \\ \max\{f_n(x)\}_{n\in\mathbb{N}}=0, & \text{otherwise } \end{cases}$$ as wanted. But I am not completely sure that my reasoning is correct, since I had never heard that the characteristic function of rationals had such representation. If correct it would also answer this question: Step functions and the characteristic function of rationals

Answer 1

I don't understand why you are making this so complicated. Just define $f_n: [a,b] \rightarrow \mathbb{R}$ by $f_n(x):= n$. Then all the $f_n$ are integrable but their max, equaling infinity on the whole interval, is not.

If this doesn't satisfy you (the max taking infinite value), and you want a supremum of countably many integrable functions which is not integrable (but still takes a finite value everywhere), just pick a sequence of step functions increasing monotonically to the function $f(x) := \frac{1}{x-a}$ for $x \in (a,b]$ and $f(a):= 0$, which is not integrable on $[a,b]$.