# Trying to understand a proof of the dominated convergence theorem

I dont understand completely a proof of the dominated convergence theorem stated in page 104 of Analysis III of Amann and Escher. I will transcribe here the proof and comment about my thoughts.

In the next: $$(X,\mathcal A,\mu)$$ is a $$\sigma$$-finite measure space, $$E$$ is a Banach space and $$\mathcal L_1(X,\mu, E)$$ is the space of Bochner $$\mu$$-integrable functions from $$X$$ to $$E$$, what is complete under the seminorm defined by $$\|f\|:=\int_X|f|\,d\mu$$.

We denote by $$|{\cdot}|$$ the norm on $$E$$, and $$\mathcal L_0(X,\mu,E)$$ is the space of $$\mu$$-measurable functions in $$E^X$$.

Let $$(f_j)$$ a sequence in $$\mathcal L_1(X,\mu, E)$$ and suppose that there exists $$g\in\mathcal L_1(X,\mu,\Bbb R)$$ such that

a) $$|f_j|\le g$$ $$\mu$$-almost everywhere for all $$j\in\Bbb N$$

Suppose also that for some $$f\in E^X$$

b) $$f_j\to f$$ $$\mu$$-almost everywhere

Then $$f$$ is $$\mu$$-integrable, $$f_j\to f$$ in $$\mathcal L_1(X,\mu,E)$$ and $$\int_Xf_j\,d\mu\to\int_X f\,d\mu$$ in $$E$$.

Proof: define $$g_j:=\sup_{k,\ell\ge j}|f_k-f_\ell|$$, then $$(g_j)\to 0$$ $$\mu$$-a.e. in $$\mathcal L_0(X,\mu,\overline{\Bbb R}^+)$$, and $$|f_k-f_\ell|\le 2g$$ $$\mu$$-a.e. for all $$k,\ell\in\Bbb N$$. Hence $$|g_j|\le 2g$$ $$\mu$$-a.e.

From a corollary of Fatou's lemma it follow that $$0\le\varlimsup_j\int_X g_j\,d\mu\le\int_X\varlimsup_j g_j\,d\mu=0$$ Therefore $$(\int_Xg_j\,d\mu)$$ is a decreasing null sequence. Then for every $$\epsilon>0$$ there is a $$N\in\Bbb N$$ such that $$\int_X|f_k-f_\ell|\,d\mu\le\int_X g_j\,d\mu<\epsilon$$ for $$k,\ell\ge j\ge N$$. Hence $$(f_j)$$ is a Cauchy sequence in $$\mathcal L_1(X,\mu,E)$$.

Up to here all is right to me, but now is the punching line what I dont follow clearly:

and the claim follow from the completeness of $$\mathcal L_1(X,\mu,E)$$ and Theorem 2.18.

Theorem 2.18 what say is that if $$(f_j)\to f$$ in the seminorm then

a) There is a subsequence $$(f_{j_k})\to f$$ $$\mu$$-a.e., and for each $$\epsilon>0$$ there is some $$A\subset X$$ with $$\mu(A)<\epsilon$$ such that $$(f_{j_k})\to f$$ uniformly in $$A^\complement$$.

b) The integral $$\int_X f_j\,d\mu$$ converges to $$\int_X f\,d\mu$$.

What I dont follow is how it follow from the completeness of the space. I mean: we knows that $$(f_j)$$ is Cauchy in the seminorm, so it converges to some value but, how we knows that it converges to $$f$$ in the seminorm?

I guess that it is because $$|f-f_j|\le\sup_{k,\ell\ge j}|f_k-f_\ell|$$, but Im not sure. However this is not related to the completeness of the space of integrable functions.

Can someone explain more clear the punching line of the proof? Thank you.

As you said, as $$(f_j)$$ is a Cauchy sequence in $$\mathcal L_1(X,\mu,E)$$ there exists some $$\phi \in \mathcal L_1(X,\mu,E)$$ such that $$f_j \to \phi$$ in the semi-norm and $$\int_X f_j d \mu \to \int_X \phi d\mu$$.
More over by the theorem mentioned there exists an extraction such that: $$f_{j_k} \to \phi \text{ a.e.}$$ but as $$f_j \to f$$ a.e. in particular you have: $$f_{j_k} \to f \text{ a.e.}$$ and by uniqueness of the limit $$f=\phi$$ a.e. which implies $$f=g$$ in $$L_1(X,\mu,E)$$.