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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.2012-12-09
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    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 $\int f = \pm \infty$.2012-12-09
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    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 $\int f = \infty$.2012-12-09
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    I agree that if $f \geq 0$, then we can talk about $\int f = \infty$. But isn't it still possible for measurable functions which alternate sign to also satisfy $\int f = \infty$?2012-12-09
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    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.2012-12-09
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    You can still make that definition, if the negative part is in $L^1$ but the positive part has infinite integral. All I'm trying to say is that the question doesn't make sense without some additional hypothesis on $f$. We are explicitly asking for $f \notin L^1$, and so you should specify what you mean by $\int f$.2012-12-09
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    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 $f_n=f\cdot\chi_{[0, 1+2+\cdots+n]}$.2012-12-09
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    @DavidMitra I believe that's an answer.2012-12-09
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    The condition in your edit is not a characterization of a function $f$ not being an element of $L_1$. A function $f$ is an element of $L_1$ if and only if $\int |f|<\infty$.2012-12-09
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    @DavidMitra: I realize that. I've edited a second time to ask the question I mean to ask. Sorry again for the confusion.2012-12-09
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    I don't think you can have it like in the edit: If $\int f_n$ do not tend to $\infty$ (say), it means that the negative parts of the $f_n$ are compensating for the positive parts, which would be in direct conflict with the condition that $\int f^-<\infty$. I can't make this a proof at the moment, but I believe David Mitra's example is the best you can hope for.2012-12-09

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