1
$\begingroup$

I was working on the proof of dominated convergence theorem, and in the proof, I saw the following argument:

By Fatou's lemma, $$\int g + \int f \leq \lim \inf \int (g + f_n) = \int g + \lim \inf \int f_n$$ and $$ \int g - \int f \leq \lim \inf \int (g - f_n) = \int g - \lim \sup \int f_n$$

So you can see that, in the second equation (in the equality part), we converted liminf to limsup. I did not understand the whole transition here, I would be very happy if anyone helps. Thanks.

$\endgroup$
1
  • 4
    $\begingroup$ Because $$\liminf \left(-\int f_n\right)=-\limsup\int f_n$$ $\endgroup$ – leo Nov 8 '12 at 4:46
0
$\begingroup$

First note that if $b_n$ is a sequence then $$\liminf_n (-b_n) = - \limsup_n b_n$$ The proof for this is relatively straight forward. $$\liminf_n (-b_n) = \lim_n \inf_{k \geq n} (-b_k) = \lim_n (-\sup_{k \geq n} (b_k)) = - \lim_n (\sup_{k \geq n} (b_k)) = - \limsup_n b_n$$ Now lets define $h_n^{(1)}(x) = g(x) + f_n(x)$ and $h_n^{(2)}(x) = g(x) - f_n(x)$.

The first assumption in LDCT is $\vert f_n(x) \vert \leq g(x)$. This implies that $h_n^{(1)}(x)$ and $h_n^{(2)}(x)$ are both non-negative over the underlying domain. Hence, by Fatou's lemma, we have that $$\int \liminf h_n^{(1)} d \mu \leq \liminf \int h_n^{(1)} d \mu$$ and $$\int \liminf h_n^{(2)} d \mu \leq \liminf \int h_n^{(2)} d \mu$$ Hence, we have that $$\int g d \mu + \int \liminf f_n d \mu \leq \int g d \mu + \liminf\int f_n d \mu$$ and $$\int g d \mu + \int \liminf (-f_n) d \mu \leq \int g d \mu + \liminf\int (-f_n) d \mu$$ The next assumption in LDCT is $\int g d \mu < \infty$. Hence, we can cancel $\int g d \mu$ from both equations to get $$\int \liminf f_n d \mu \leq \liminf\int f_n d \mu$$ and $$\int \liminf (-f_n) d \mu \leq \liminf\int (-f_n) d \mu$$ Now since $$\liminf_n (-b_n) = - \limsup_n b_n$$ the second equation can be written as $$\int -\limsup (f_n) d \mu \leq -\limsup\int f_n d \mu$$ which can be rewritten as $$\int \limsup (f_n) d \mu \geq \limsup\int f_n d \mu$$ The last assumption in LDCT is that $f_n \to f$. Hence, we have that $\limsup (f_n) = f = \liminf (f_n)$. Hence, we get that $$\limsup\int f_n d \mu \leq \int f d \mu \leq \liminf\int f_n d \mu$$ But we know that for any sequence $\{a_n\}$, $$\liminf_n a_n \leq \limsup_n a_n$$ and equality holds only when limit exists. Hence, putting all this together, you get that $$\lim_n \int f_n d \mu = \int f d \mu$$

$\endgroup$

Your Answer

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