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