I'm doing an exercise in Lebesgue integral.
Definitions probably used:
Some of them can be acquired here: Questions of an exercise in Lebesgue integral.
Definition of Convergence in measure: Suppose $f(x), f_1(x), f_2(x),..., f_k(x), ...$ are Lebesgue measurable functions that are finite almost everywhere ($m(\{x \in E: |f_k(x)| = +\infty\}) = 0$). $\{f_k(x)\}$ converge in measure to $f(x)$ if for $\forall \epsilon > 0,$ $\lim_{k \to \infty} m(\{ x\in E: |f_k(x) - f(x)| \ge \epsilon\}) = 0$.
Exercise3:
Plese note it is based on Lebesgue integral.
Suppose $0 \le f_1(x) \le f_2(x) \le ... \le f_k(x) \le ...(x \in E)$ and $f_k(x)$ converges in measurable to $f(x)$, then show $$\lim_{k \to \infty} \int_{E} f_k(x) dx = \int_{E} f(x) dx.$$
My trial and trouble:
There is no solution for this exercise. I tried to use Riesz's theorem of convergence in measure that is,
suppose $\{f_k(x)\}$ converge in measure to $f(x)$, then there exist a subsequence of $\{f_k(x)\}$, named $\{f_{k_i}(x)\}$ such that $\lim f_{k_i}(x) = f(x),$ almost everywhere, $ x \in E.$
I can prove $\lim_{i \to \infty} \int_{E} f_{k_i}(x) dx = \int_{E} f(x) dx.$ However, I don't know how to show $\lim_{k \to \infty} \int_{E} f_k(x) dx = \lim_{i \to \infty} \int_{E} f_{k_i}(x) dx.$ The trouble is I can take $\lim$ into $\int_{E} f_{k_i}(x)$ by monotone convergence theorem that is $$\lim_{i \to \infty} \int_{E} f_{k_i}(x) dx = \int_{E} \lim_{i \to \infty} f_{k_i}(x) dx.$$ But I can't do that for $\lim_{k \to \infty} \int_{E} f_k(x) dx$ coz $\lim_{k \to \infty} f_k(x)$ may not exist.
For each $x\in E$ we have $\lim_{i\to\infty } f_{k_i} (x) =\lim_{k\to\infty} f_k (x)$ since $(f_n)$ is monotone. Analogously for the monotone sequence $a_k =\int_E f_k (t) dt $ we get $\lim_{i\to\infty} a_{k_i} =\lim_{k\to \infty } a_k .$