Proving $f$ is not Riemann integrable but is Lesbesgue with definition involving simple step functions

Question

I am familiar with the proof of showing that $f:[0,1]\to\mathbb{R}$, where $$f(x)=\begin{cases} 0 & \text{if} \, \, x\in\mathbb{R}\setminus\mathbb{Q} \\ 1 & \text{if} \, \, x \in \mathbb{Q} \end{cases}$$ is not Riemann integrable with the traditional definition of Riemann integration with Darboux sums and partitions of intervals, etc. However, I want to show the same result with an alternative definition. Additionally, I want to show that it is Lesbesgue integral using a particular definition as well.

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Definition: A function $f$ is said to be Riemann integrable if for all $\epsilon>0$ there exists simple step functions $g,h$ such that $g \leq f \leq h$ and $\int h - g < \epsilon.$

Definition: By a simple step function we mean a finite linear combination of characteristic functions on semi-open intervals, i.e. if $s$ is a simple function then: $$s:=\lambda_1 \chi_{I_{1}} + \dots + \lambda_n \chi_{I_{n}}$$ where $\lambda_j$ are scalars and $I_j$ are semi-open intervals for $1 \leq j \leq n$.

Definition: A real valued function $f$ defined on $\mathbb{R}^n$ is said to be Lesbesgue integrable if there exists a sequence of simple step functions, $(f_n)$ such that the following two conditions are satisfied:

a) $\quad \displaystyle\sum_{n=1}^{\infty} \int |f_n| < \infty$

b) $\quad f(x)=\displaystyle\sum_{n=1}^\infty f_n(x) \quad \forall x \in\mathbb{R}^n$ such that $\displaystyle\sum_{n=1}^{\infty} |f_n| < \infty.$

Then, the integral of $f$ is defined to be: $$\int f = \sum_{n=1}^{\infty} \int f_n.$$

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I'm not sure how to show the result with this definition, but I'm thinking of going about it by contradiction, suppose $f$ is Riemann integrable, then there are two simple step functions $g$ and $h$ such that $g \leq f \leq h$. Then, maybe we could choose a sequence $(g_n)$ and $(h_n)$ such that $g_n \to g$ and take subsequences such that each $g_n \in \mathbb{Q}$ for all $n$, and similarly $h_n \to h$ and take subsequences such that each $h_n \in \mathbb{R}\setminus\mathbb{Q}$, for all $n$, so $\int h - g = 1$, which is not less than $\epsilon$ for all $\epsilon>0$. This doesn't quite feel right though.

Then, as far as showing that it is Lesbesgue integable with this definition, I am thinking about enumerating a sequence of rational numbers $\left\{ q_1 , q_2 , \dots \right\}$ such that $f_n = q_n$ and using that but as my simple functions, but I'm having trouble formalizing it with this definition.

Answer 1

As to the first part, since any interval contains both rational and irrational points, a simple step function $\ge f$ must be $\ge 1$ everywhere on $I=[0,1]$, and a simple step function $\le f$ must be $\le 0$ everywhere on $I.$

I thought I had the second part solved, but just as I finished typing my answer, I saw that it was wrong or at least incomplete. I'm going to post it anyway, because it's very close, and I think you can finish it.

It's enough to find for each $q\in I\cap \mathbb Q,$ a sequence of simple step functions $f_{q,n}$ such that $$\sum_{n=1}^{\infty}{f_{q,n}(x)}= \begin{cases} \infty, &x=q\\ 0,&x\in I\setminus\mathbb Q \end{cases} $$ I haven't specified the value of the sum at rationals in $I$ other than $q$. It's going to be $\le 0$ at such points. The difficulty I'm having is arranging that for any $x\in I$ there are only finitely many $q$ such that $\exists n, f_{q,n}(x)< 0.$ It will be clear from the construction that for any $q$ there is at most one such $n$.

Since the countable union of countable sets is countable, we can arrange the $f_n,q$ into a sequence $f_n$ with the required property: $$\sum_{n=1}^{\infty}{f_n(x)}= \begin{cases} \infty, &x\in I\cap \mathbb Q\\ 0,&x\in I\setminus\mathbb Q \end{cases} $$ To construct $f_{n,q}$ for a given $q\in I\cap\mathbb Q,$ choose a strictly decreasing sequence of rational numbers $\varepsilon_n \rightarrow 0$ such that $[q,q+\varepsilon_n)\subset I, n=1,2,3\dots.$ Now for $n=1,2,3\dots,$ define $$f_{q,2n-1}=n\chi_{[q,q+\varepsilon_n)},\\ f_{q,2n}=-n\chi_{(q,q+\varepsilon_n]}$$

If it weren't for the "semi-open interval" requirement, we could make the interval in the definition of $f_{q,2n}$ equal to $(q,q+\varepsilon_n)$ and we'd be done. Now it seems as if we have to somehow control the $\varepsilon_n$ (which depend on $q,$ remember) so that no point occurs as an upper endpoint of the interval more than finitely many times.

This seems straightforward, but I haven't worked out all the details. Arrange the elements of $I\cap\mathbb Q$ in sequence: $q_1, q_2, \dots.$ There's no problem finding suitable $\varepsilon_n$ for $q_1$. For $q_2$ choose a sequence $\varepsilon_n'\rightarrow 0.$ Since $q_1+\varepsilon_n\rightarrow q_1, q_2+\varepsilon_n'\rightarrow q_2,$ only finitely many of the $q_2+\varepsilon_n'$ can be in the sequence $q_1+\varepsilon_n,$ and we can simply delete the corresponding $\varepsilon_n'$ from the sequence. A similar arguments works for $q_k,$ because we have at most finitely many $\varepsilon$ to discard from each of the $k-1$ preceding sequences.

I think this is a sketch of a valid proof, but it needs to be cleaned up, obviously. The only other detail is that the argument doesn't work at $q=1,$ where there is no interval $[q,q+\varepsilon)\subset I,$ but a symmetric argument works.