Want to prove that limit of integrals diverges

[jostpuur]

Suppose we know that f_n\to f when n\to\infty (in some sense), and that

<br /> \int f(x)dx<br />

diverges. Can it be deduced that

<br /> \lim_{n\to\infty} \int f_n(x)dx<br />

diverges too?

 

[elibj123]

If the convergence of the sequence is uniform on the interval of integration then yes

 

[jostpuur]

I see now that this seems to follow from the Fatou's lemma

I thought first it wouldn't be relevant, because I wanted to consider complex functions, but if f is complex, and the integral diverges, then it means that at least one positive or negative part, of real or imaginary part, diverges.

Then, for example if the (\textrm{Re}(f))^+ diverges, then

<br /> \lim_{n\to\infty} \int \textrm{Re}(f_n(x))^+ dx = \underset{n\to\infty}{\textrm{lim inf}} \int \textrm{Re}(f_n(x))^+ dx \geq \int \textrm{Re}(f(x))^+ dx = \infty<br />

I guess that's it.

 

[jostpuur]

But I don't think that my original problem would be solved yet, because my original problem looks more like this:

Suppose X is some measure space such that \mu(X)=\infty and X_1\subset X_2\subset\cdots is some subset sequence such that \mu(X_k)&lt;\infty for all k and such that

<br /> \bigcup_{k=1}^{\infty} X_k = X<br />

No I know that point wisely f_n(x)\to f(x), and that

<br /> \int\limits_X |f(x)| d\mu(x) =\infty<br />

and I want to prove that

<br /> \lim_{n\to\infty} \lim_{k\to\infty} \int\limits_{X_k} f_n(x) d\mu(x)<br />

diverges.

So is it certain, that the integrals of f_n could not start converging for cancellation reasons?

 

[jostpuur]

I've been "writing while thinking" so it could be that my posts are little confused.

 

[quasar987]

First, since for each n, the map

\nu(A)=\int_Af_nd\mu

is a measure, we have by continuity, that

<br /> \lim_{n\to\infty} \lim_{k\to\infty} \int\limits_{X_k} f_n(x) d\mu(x)=\lim_{n\to\infty}\int_Xf_nd\mu<br />

So your question amounts to: is there a sequence of integrable functions f_n that converge pointwise to a non integrable function f but such that \lim_n\int_Xf_n converges.

The answer is provided by the old example of X=R, f(x)=sin(x)/x. It is fairly easy to show that \int_{\mathbb{R}}|f|=+\infty by finding an appropriate sequence of step function bounded above by |f| and whose area makes up a diverging series.

But it is known that the sequence of (integrable) functions

f_n(x):=\chi_{[-n,n]}\frac{\sin(x)}{x}

(which converge pointwise to f) are such that

\lim_{n\to\infty}\int_{\mathbb{R}}f_n=\pi

(more difficult).

So, here is your counter-example.