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.

  • $\begingroup$ Do you want to assume $f$ is nonnegative, then? Else the symbol $\int f$ has no meaning. $\endgroup$ – Zach L. Dec 9 '12 at 2:51
  • $\begingroup$ 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 $\int f = \pm \infty$. $\endgroup$ – Jesse Madnick Dec 9 '12 at 2:58
  • $\begingroup$ 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 $\int f = \infty$. $\endgroup$ – Zach L. Dec 9 '12 at 3:00
  • 2
    $\begingroup$ 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. $\endgroup$ – Jose27 Dec 9 '12 at 3:04
  • 1
    $\begingroup$ 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 $f_n=f\cdot\chi_{[0, 1+2+\cdots+n]}$. $\endgroup$ – 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 $f_n(x)= {1\over x}\cdot\chi_{[1,n]} + {-1\over n}\cdot\chi_{(n, n+ n\ln n )}$.


$\ \ \ 1)\ \int_{\Bbb R} f=\infty$,

$\ \ \ 2)\ (f_n)$ converges to $f$ uniformly, since $\Vert f_n-f\Vert_\infty=2/n$,


$\ \ \ 3)\ $for each $n$ we have $\int_{\Bbb R} f_n=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.