I have a question about a proof of the dominating convergence theorem, with weak requirements. Before I show the proof from the book, note that in my book you are allowed to integrate functions that are not defined everywhere, but almost everywhere, the make this definition, it may be relevant to know:
Now comes the part I do not fully understand. The proof deals with both the real and complex case, but I am only interested in the real case(green). I have outlined what I don't udnerstand(red). They say that for each function they can just assume that $|f_n|<g$ everywhere, not almost everywhere:
Why can they just assume this?I think it follows because if this doesn't hold we may use some kind of trick to replace these functions with functions that are smaller than g everywhere. But I don't fully understand precisely the details and how I am supposed to do this? Can you please help me?
Because for practical purposes null sets are irrelevant. Formally, let $A_n=\{x: |f_n(x)|> g(x)\}$.By assumption $\mu(A_n)=0$ for each $n$. Now, define $\overline{f_n}(x)=\begin{cases} f_n(x) \ \rm{if}\ x\notin A_n\\ 0 \ \rm{otherwise} \end{cases}$. Since $f_n$ and $\overline{f}_n$ are equal almost everywhere their integrals are the same. So the right hand side limit is the same for both sequences. On the other hand, let $\overline{f}(x)=\begin{cases} \lim_{n\rightarrow\infty}f_n(x) \ \rm{if\ the\ limit\ exists}\ \\ 0 \ \rm{otherwise} \end{cases}$.
Observe that $\{x:f(x)\ne \overline{ f}(x)\}\subseteq \bigcup_{n=1}^\infty A_n$, since a countable union of measure zero sets has measure zero it follows that $\overline{f}$ and $f$ are equal almost everywhere. Thus the left hand side integral are also the same.