# 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 $\int f_n \not \to \int f$

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.

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Do you want to assume $f$ is nonnegative, then? Else the symbol $\int f$ has no meaning. –  Zach L. Dec 9 '12 at 2:51
Well, $\int f$ is meaningful if $\int f = \pm \infty$. In fact, that's really what I'm after: $\int f_n \not \to \int f$, where $\int f = \pm \infty$. –  Jesse Madnick Dec 9 '12 at 2:58
Right, but what is the integral of something like $f(x) = \sin(x)$? The positive and negative parts are both infinite. On the other hand, if the function is positive, then it makes sense to define $\int f = \infty$. –  Zach L. Dec 9 '12 at 3:00
If $f\geq0$ this can't happen: $f\notin L^1$ implies that the integral is infinite. On the other hand if the limit were finite (in fact it suffices to assume the $\liminf$ to be finite) Fatou's lemma gives $f\in L^1$, a contradiction. @ZachL. You can still define the integral if either $f^+$ or $f^-$ are finite. Of course your example doesn't satisfy this. –  Jose27 Dec 9 '12 at 3:04
Let $f$ have the value $1$ on $[0,1)$, the value $-1/2$ on the interval $[1,3)$, the value $1/3$ on the interval $[3,6)$, $\ldots$ and let $f_n=f\cdot\chi_{[0, 1+2+\cdots+n]}$. –  David Mitra Dec 9 '12 at 3:17

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 $\int_{\Bbb R} f_n=0$.

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Thank you! ${}{}{}$ –  Jesse Madnick Dec 9 '12 at 5:19