1
$\begingroup$

I was going through the proof of the Dominated Convergence Theorem.

Now if we have that $(f_n)$ is a sequence of measurable functions such that $\lvert f_n\rvert\le g$ for all $n$ where $g$ is integrable on $\Bbb R$. And if $f = \lim_n f_n$ almost everwhere.

We can show that $(g+f_n)$ and $(g-f_n)$ are sequences of non-negative measurable functions.

How can we show this for $(g-f_n)$? We do get that $f_n\le g$, but what if $f_n(x)$ is infinite for some $x$?

Then by Fatou's lemma, we have that $\int\liminf(g+f_n)dx\le\liminf\int(g+f_n)dx$. Now from here, we can obtain that

$$\int(g+f)dx \le \int gdx+\liminf \int f_ndx$$

How? How do we get the left hand side in this? I can show that $\int\liminf f_ndx = \int fdx$, but how to prove that $\liminf f_n$ is integrable to prove the former? Also I know that $\liminf(g+f_n) \geq\liminf g+\liminf(f_n)$... but how do we get the left hand side? Can I integrate it throughout but why are all these limit inferior integrable?

$\endgroup$
3
  • $\begingroup$ you are asking about why $\liminf_n (c+x_n)=c+\liminf_n x_n$ for some constant $c$? $\endgroup$
    – Masacroso
    Jun 10 '20 at 19:25
  • 1
    $\begingroup$ @Masacroso: for the second time. $\endgroup$ Jun 10 '20 at 21:13
  • $\begingroup$ @MartinArgerami..no..that doubt is clear..I even accepted that answer $\endgroup$
    – Gitika
    Jun 11 '20 at 6:23
0
$\begingroup$

I assume there are three major doubts. $\\$

1) What happens when $f_n(x)$ is infinite for some $x$? $\\$

2) Why $ \liminf \int (g+f_n)d\mu=\int g d\mu + \liminf f_nd\mu ? \\$

3) Why $ \liminf f_n $ is integrable? $\\$

Let's consider them one by one. $\\$

1) Note that $f_n$ is dominated by $g$ which is in $L^1$ implying that $f_n$ is in $L^1$, hence, is finite a.e. So, even if $f_n$ is infinite at some point, it doesn't matter because outside a measure zero set, $f_n$ is finite for every $n$ and measure zero sets are killed during integration. $\\$

2) Integration and limit are linear. Since $\liminf$ is actually limit in this case, so, $ \liminf \int (g+f_n)d\mu=\liminf \int g d\mu + \liminf f_nd\mu \\$ . Since first integral with $g$ is independent of $n$. Hence, $\liminf \int g d\mu= \int g d\mu \\$

3) First note that $\liminf f_n$ is measurable because each $f_n$ is measurable and since each $f_n$ is bounded by $g$, hence, $\liminf f_n = \sup_{n\in \mathbb{N}} \inf_{m\geq n} (f_m) \leq g $ and therefore is integratable as $g$ is integrable.

$\endgroup$
6
  • $\begingroup$ Note that liminf is not additive in general. It is in this case because it is actually a limit. $\endgroup$ Jun 10 '20 at 21:13
  • $\begingroup$ why is $\liminf$ the limit in that case?.how to know that $\int$(g+fn) is convergent? $\endgroup$
    – Gitika
    Jun 11 '20 at 7:40
  • $\begingroup$ The fact that $\int$fn is convergent has been proven in the result..this what I am asking is an intermediate step $\endgroup$
    – Gitika
    Jun 11 '20 at 7:41
  • $\begingroup$ And also.can we also prove that liminf(fn+g) is integrable? $\endgroup$
    – Gitika
    Jun 11 '20 at 7:42
  • $\begingroup$ By proving that liminf(g+fn) is bounded by 2g? $\endgroup$
    – Gitika
    Jun 11 '20 at 7:50

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.