Pointwise limit of integrable function Question:

Show that the pointwise limit of integrable functions is not necessarily integrable.

I am stuck on this question. Here is what I know.
Let $(f_n)^{\infty}_{n=1}$ be a series of integrable functions, and let 
$$\lim_{n \to \infty} f_n(x)=Z$$
I need to show that $Z$ is not necessarily integrable. Should I be looking for a specific example? A function that is integrable, but as $n \to \infty $ the function is no longer integrable.
 A: Yes, a specific example is what you need, and you might start by looking at a well-known function which is not integrable (for example $f(x)=1/x$) and then find some integrable functions which converge to $f$ - e.g. consider the functions $f_n(x) = \max(f(x),n)$. I will let you check the details ...
A: Here is another natural example on $\mathbb{R}$ with the Lebesgue measure, which works also on $\mathbb{Z}$ with the counting measure:
$$
f_n=1_{[-n,n]}
$$
Use the Monotone Convergence Theorem.
In the same spirit, we can do even worse. Take
$$
g_n=-1_{[-n,0)}+1_{(0,n]}.
$$
Then all the $g_n$'s have $0$ integral. So $\lim
_n \int g_n=0$ exists. Yet the pointwise limit is not integrable. Again neither on $\mathbb{R}$ nor on $\mathbb{Z}$.
A: Just in case your question is about Riemann integrability, I will provide a sequence of Riemann integrable functions $f_n\colon[0,1]\to\mathbb{R}$ such that $\lim_{n\to\infty}f_n(x)=f(x)$ exists for all $x\in[0,1]$ but $f$ is not integrable in the sense of Riemann.
Order the rationals in $[0,1]$:
$$
[0,1]\cap\mathbb{Q}=\{r_n\}_{n=1}^\infty.
$$
Let
$$
\phi_n(x)=\begin{cases}
1 &\text{if }x=r_n,\\
0 &\text{otherwise}
\end{cases}
\quad\text{and}\quad f_n(x)=\sum_{k=1}^n\phi_k(x).
$$
Then $f_n$ is Riemann integrable, since it has a finite number of discontinuities ($r_1,\dots,r_n$), but
$$
f(x)=\lim_{n\to\infty}f_n(x)=\sum_{k=1}^\infty\phi_k(x)=\begin{cases}
1 &\text{if }x\text{ is rational}\\
0 &\text{if }x\text{ is irrational}
\end{cases}
$$
is not Riemann integrable (althoug it is Lebesgue integrable).
A: Check this,
$$ \int_{0}^{1}x^{-1+\frac{1}{n}}dx. $$
Note that, the functions $x^{1+1/n}$ are integrable as improper integrals or in the extended sense of Riemann and Lebesgue. 
