5-5

Question 5.5. Use Egorov's theorem to prove the bounded convergence theorem.
Solution.
Suppose \left\{f_{k}\right\} is a sequence of measurable functions such that f_{k} \rightarrow f  a.e. in  E , where  |E|<+\infty , and so that  \left|f_{k}(x)\right| \leq M  for all  x \in E . Given  \varepsilon>0, by Egorov's theorem there exists a closed set F \subseteq E  such that |E - F|<\varepsilon  and on which f_{k} \rightarrow f  uniformly. That is, there exists K>0 such that every k for k>K,\left|f_{k}-f\right|<\varepsilon  on  F.
Then we can have 
\left|\int_{E} f_{k}-f\right| \leq \int_{E - F}\left|f_{k}-f\right|+\int_{F}\left|f_{k}-f\right| \leq \int_{E - F} 2 M+\int_{F} \varepsilon \leq 2 M|E - F|+\varepsilon|E|<(2 M+|E|) \varepsilon
.
Since |E|<+\infty, we conclude that for every  \varepsilon>0, there exists K>0 such that \left|\int_{E} f_{k}-f\right|< c \varepsilon for every k>K, where c is a constant, so  \int_{E} f_{k} \rightarrow \int_{E} f  as  k \rightarrow \infty.