# How Lebesgue integration solved the problem of a function being integrable but its limit is not integrable?

My professor gave us the following form of Dirichlet function as an example of the problems we faced in Riemann integration:

$$\{r_{n}\}$$ enumeration $$\mathbb{Q} \cap [0,1]$$

$$f_{n}(x) = \begin{cases} 1 & \quad x \in \{r_{1}, ... , r_{n}\} \\ 0 & \quad \text{otherwise}. \end{cases}$$

And he said that: each $$f_{n}$$ is integrable but its limit is not integrable.

My questions are:

1- why each $$f_{n}$$ is integrable but its limit is not integrable?

2- How did Lebesgue integration solve this problem?

Could anyone help me understand answers to these questions, please?

• 1. follows from looking at the Darboux sums. Commented Jun 18, 2020 at 15:30
• A typical approach to Lebesgue integral is to define the integral of a limit of sequence of functions as limit of integrals of these functions (this involves lot of technical details which I avoid here). So this is a feature by design. Commented Jun 18, 2020 at 15:54
• In the current case we have $\int f_n=0$ and thus we define the Lebesgue integral $\int f=\lim\int f_n=0$. Commented Jun 18, 2020 at 15:54
• 2. If you study Lebesgue measure & integration you will understand why, the TL;DR is that Lebesque integration (well, measure really) allows the domain to be partitioned in a much more refined manner than Riemann integration. A finite number of non trivial intervals must contain point where $f$ is one and zero, whereas using a partition of the form $\mathbb{Q} \cap [0,1]$ and $[0,1]\setminus \mathbb{Q}$ allows one to deduce that the integral is zero. Commented Jun 18, 2020 at 15:55
• No, my comment is related to definition of Lebesgue integrals. A specific version is: let $f_n$ be an increasing sequence of step functions on interval $I$ such that $\lim \int_{I} f_n$ exists. Then $f_n$ converges almost everywhere in $I$ to some function $f$ and then we define $\int_{I} f=\lim\int_{I} f_n$. Difference of two functions like $f$ gives the class of Lebesgue integrable functions on $I$. Commented Jun 19, 2020 at 0:01

1. Each $$f_n$$ is integrable because you can check that each lower sum is equal to $$0$$ and that the upper sums can take (positive) values as small as you want. Therefore, the (Riemann) integral of $$f$$ is $$0$$. However, if $$f$$ is the limit, then every lower sum is $$0$$ and every upper sum is $$1$$. Therefore, $$f$$ is not Riemann-integrable.
2. In the case of Lebesgue integration, $$f$$ is integrable, and its integral is $$0$$.
• Not at all. I am just using that fact that if I have two functions $f,g\colon[a,b]\longrightarrow\Bbb R$ such that $\{x\in[a,b]\mid f(x)\ne g(x)\}$ has Lebesgue measure $0$ and that $g$ is Lebesgue-integrable, then $f$ is Lebesgue-integrable too and the integrals are equal. Apply this to my function $f$ and take $g$ equal to the null function. Commented Jun 18, 2020 at 15:39
• No. I am using the fact that the set $\mathbb Q\cap[0,1]$ has Lebesgue measure $0$. Commented Jun 21, 2020 at 14:37