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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 $a<b \in \mathbb{R}, \mu(a,b)=b-a$.

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.

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

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