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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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