The following is a proof the dominated convergence theorem from Folland

Proof. $f$ is measurable (perhaps after redefinition on a null set) by Propositions 2.11 and 2.22, and since $\left |f \right | \leq g$ a.e., we have $f\in L^{1}$. By taking real and imaginary parts it suffices to assume that $f_n$ and $f$ are real-valued, in which case we have $g+f_n \geq 0$ a.e. and $g-f_n \geq 0$ a.e. Thus by Fatou's lemma,

$\int g + \int f \leq \lim \inf \int (g+f_n) = \int g + \lim \inf \int f_n$

$\int g -\int f \leq \lim \inf \int (g-f_n) = \int g - \lim \sup \int f_n$

Where do the equalities from both lines come from? Is it true that in general, $\lim \inf ({a_n + b_n}) = \lim \inf a_n + \lim \inf b_n$?

Where does the $\lim \sup$ come from the second line?


The first fact is true in general.

about the second line: $$\liminf\int (g-f_n)=\liminf(\int g-\int f_n)$$ So the $\liminf$ converges to the smallest difference between the values of $\int g$ and $\int f_n$ and since $|f_n|\leq g\;\;\forall n\in \mathbb N$ and $g$ does not depend on $n$, the difference is "minimal" if $\int f_n$ is "maximal" and that happens for the $\limsup\int f_n$.

  • $\begingroup$ I think it follows because in general because $\lim \inf (-c_n) = -\lim \sup c_n$. It doesn't have anything to do with $|f_n| \le g$ $\endgroup$ – user135872 Mar 16 '14 at 21:35
  • $\begingroup$ yes you are absolutely right, sorry for that. And i have to apologize since the first fact is also not true in general. But it holds that $\limsup(a_n+b_n)\leq\limsup a_n + \limsup b_n$ and $\liminf(a_n+b_n)\geq\liminf a_n +\liminf b_n$. Sorry for the bad answer! $\endgroup$ – Thorben Mar 16 '14 at 22:35
  • $\begingroup$ You can edit the bad answer into a good answer... which will likely get upvoted. $\endgroup$ – user127096 Mar 29 '14 at 2:14

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