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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 ($g+f_n$) is a sequence of non-negative measurable functions.

Then by Fatou's lemma, we have that $\int$liminf($g+f_n$)$d$$x$ $\le$ liminf$\int$($g+f_n$)$d$$x$.

Now from here, we can obtain that

$\int$($g+f$)$d$$x$ $\le$ $\int$$gdx$+liminf $\int$$f_ndx$.How?

Please explain this last step.I know that since both are integrable, the integral can be seperated..but how is liminf seperated in the right-hand side?

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1 Answer 1

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Inside the liminf, $\int g$ is just a number. So what it says is that $$ \liminf_n(c+f_n)=c+\liminf _nf_n, $$ where $c=\int g$.

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  • $\begingroup$ So..is that always true? ..liminf(c+an) = c+liminfan? $\endgroup$
    – Gitika
    Jun 10, 2020 at 5:19
  • $\begingroup$ Yes. How could translating a set by a fixed distance could change it in any other way? If you move a set 10 units to the right, the infimum will increase by 10. $\endgroup$ Jun 10, 2020 at 5:21
  • $\begingroup$ Are you sure $\int$g is a number? $\endgroup$
    – Gitika
    Jun 10, 2020 at 5:39
  • $\begingroup$ for example..what if g(x)=x? $\endgroup$
    – Gitika
    Jun 10, 2020 at 5:40
  • $\begingroup$ I can see it is independent of n..it's a constant sequence $\endgroup$
    – Gitika
    Jun 10, 2020 at 5:41

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