TITLE: Example of a sequence $f_n \in L^1(\mathbb{R})$ with $f_n \to f$ uniformly, but such that $f \not \in L^1(\mathbb{R})$ and $\int f_n \not \to \int f$
QUESTION [5 upvotes]: For practice, I'm working through some of the exercises in Folland's "Real Analysis: Modern Techinques and Their Applications."
In Chapter 2, Exercise 19, Folland asks for sequences of functions $f_n \in L^1(\mathbb{R})$ with $f_n \to f$ uniformly, but such that one of the conclusions $f \in L^1(\mathbb{R})$ or $\int f_n \to \int f$ fails.  I can find examples of each conclusion failing, but cannot seem to find a single example where both conclusions fail in the following way:

What is an example of a sequence of functions $f_n \in L^1(\mathbb{R})$ with $f_n \to f$ uniformly, but such that $\int f = \pm \infty$ (by which I mean that exactly one of $\int f^+$ or $\int f^-$ is $+\infty$) and also $\int f_n \not \to \int f$?

My examples:
(1) $f_n(x) = \frac{1}{x}\chi_{(1,n)}(x)$.  The uniform limit $f(x) = \frac{1}{x}\chi_{(1,\infty)}(x)$ has $\int f = +\infty$.  However, we also have $\int f_n = \log(n) \to \infty = \int f$.
(2) $f_n(x) = \frac{1}{n}\chi_{(0,n)}(x)$.  We have $\int f_n = 1 \not \to \int f = 0$, but now $\int f = 0 \neq \pm \infty$.
I feel like I'm missing something very obvious here.  Thanks for your help.

REPLY [7 votes]: You just need to modify your first example:
Take $f(x)={1\over x}\cdot\chi_{[1,\infty)}$ and for $n$ a positive integer, define
$f_n(x)= {1\over x}\cdot\chi_{[1,n]}  + {-1\over n}\cdot\chi_{(n, n+ n\ln n )}$.
Then 
$\ \ \ 1)\  \int_{\Bbb R} f=\infty$, 
$\ \ \ 2)\ (f_n)$ converges to $f$ uniformly, since $\Vert f_n-f\Vert_\infty=2/n$, 
and 
$\ \ \ 3)\ $for each $n$  we have $\int_{\Bbb R} f_n=0$.