Understanding the proof of the Dominated Convergence Theorem

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?

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

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