# Find a example where $\{\phi_{n}(x)\}$ are simples and Lebesgue-integrables but limit of integration doesn't exist

Definition: A function, $$f: X\rightarrow \mathbb{R}$$ is simple, if $$f$$ is Lebesgue-measure and takes a number to the most numberable of different values.

Definition: A simple function, $$f,$$ is Lebesgue-integrable if $$\sum_k y_k \mu(A_k)$$
absolutely converges, where $$A_{k}=\{x \in X: f(x)=y_{k} \}$$ and $$y_{k}$$ are different values of $$f$$

Observation: Lebesgue-measure for a set $$A$$ is non-negative.

Problem. Find a example where $$\{\phi_n(x)\}$$ are simple and Lebesgue-integrable but limit of integration doesn't exist.

My attempt:

Take $$\mathbb{R}$$ with measure of intervals; for $$a.

Let $$\{\phi_{n}(x)\}$$ defined by

$$\phi_{n}(x) = \begin{cases} nx &\quad\text{if }x \in \left[-n^{\frac{1}{2}},\frac{1}{n^3} \right] \\ 0 &\quad\text{otherwise.} \\ \end{cases}$$

When $$\phi_{n}(x)$$ takes $$0$$, $$\phi_{n}(x)$$ is measurable and Lebesgue-integrable because $$\sum_{k}|0|\mu(\mathbb{R-[-n^{\frac{1}{2}},\frac{1}{n^3}]})=0$$. And when takes $$nx$$, $$\phi_{n}(x)$$ is measurable for each $$n \in \mathbb{N}$$, fixed, because $$A_{n}=\{x \in \mathbb{R}: f(x)=nx \}=[-n^{\frac{1}{2}},\frac{1}{n^3}]$$ and $$\mu[-n^{\frac{1}{2}},\frac{1}{n^3}]=\frac{1}{n^3}+n^{\frac{1}{2}} > 0$$. $$\phi_n(x)$$ is Lebesgue-integrable because $$\sum_n|nx|\mu[-n^{\frac{1}{2}},\frac{1}{n^3}]=\sum _{n}n|x|(\frac{1}{n^3} + n^{\frac{1}{2}}) \leq \sum_{n}n \cdot \frac{1}{n^3}(\frac{1}{n^3} + n^{\frac{1}{2}}) = \sum_n \frac{1}{n^6}+\frac{1}{n^{\frac{3}{2}}}$$, where that sum converges then the original sum absolutely converges.

We can see that $$\{\phi_{n}(x)\}$$ converges uniformly, using the definition on sequences uniformly, to $$\phi(x)=0$$ $$\forall x \in \mathbb{R}$$. But,

$$\int_{\mathbb{R}}\phi_{n}(x) \, d\mu(x) = \int_{-\infty}^{+\infty} \phi_n(x) \, dx = \int_{-n^\frac{1}{2}}^{\frac{1}{n^3}} nx \, dx=[n\frac{x^2}{2}]_{-n^{\frac{1}{2}}}^{\frac{1}{n^3}}=\frac{1}{2n^5}-\frac{n^2}{2}$$

we can see that $$\lim_{n \rightarrow \infty}\int_{\mathbb{R}}\phi_n(x) \,d\mu(x)=\lim_{n\rightarrow \infty}\left(\frac{1}{2n^5}-\frac{n^2}{2} \right) = -\infty$$, so the limit doesn't exist.

But my teacher told me that $$\phi_n(x)$$ is not simple because it has a continuous set of values. Could you help me to give a better example?, please.

• Why not just take $\phi_n=n\cdot 1_{[0,1]}$, where the latter is the indicator function of $[0,1]$? – copper.hat Mar 25 at 17:14
• Then take $n \cdot 1_{[-n,n]}$. Simple, integrable, limit of integral is $\infty$. – copper.hat Mar 25 at 17:21
• but when I see if this function is Lebesgue integrable, $\sum_{n} n$ is not absolutely converges – PSW Mar 25 at 17:26
• I don't understand what you are trying to say. What has the sum got to do with the above? – copper.hat Mar 25 at 17:28
• A simple function, $f,$ is Lebesgue-integrable if $\sum_k y_k \mu(A_k)$ absolutely converges, where $A_{k}=\{x \in X: f(x)=y_{k} \}$ and $y_{k}$ are different values of $f$ – PSW Mar 25 at 17:29

## 1 Answer

Edit: I apparently missed that your actual question is "what are some better examples", not "why is $$\phi_n$$ not simple". So please just consider this a comment that was too long. Edit 2: apparently also I was channelling Jordan instead of Lebesgue - modified to match the Lebesgue definition of a simple function.

Your problem is here:

Definition: A function, $$f: X\rightarrow \mathbb{R}$$ is simple, if $$f$$ is Lebesgue-measure and takes a number to the most numberable of different values.

Perhaos this is just a really poor translation, but that "definition" doesn't make any sense in English.

The proper definition of a simple function is: "$$f : X \to \Bbb R$$ is simple if $$f$$ is Lebesgue-measurable, and takes on only a countable number of values."

That means $$f(X)$$ is a countable set.

For your functions, $$\phi_n$$ takes on every value between $$-\sqrt n$$ and $$n^{-2}$$, which is an uncountable set. So your functions are not simple.

Copper hat has supplied a couple better examples. (The sum for Lebesgue integrability is $$0 \cdot \mu((-\infty, -n)\cup(n,\infty)) + n \times \mu([-n,n]) = 0 + 2n^2$$. "$$n$$" is a constant for this calculation, not an index of the summation.)