This exercise is the other one that I have some trouble with.
Some usual definitions can be acquired here: Questions of an exercise in Lebesgue integral, (obviously, that was the first one which I got stuck by).
Another definition probably used, Dominated Convergence Theorem: Suppose $f_k \in L(E) (k=1,2,3,...)$ where $L(E)$ means a set of measurable functions defined on $E \subset \mathbb R^n$ while each function $f$ in it satisfies $||f||_{1} = \int_{E}|f(x)|dx < \infty$(Note: it is a Lebesgue integral). And $$\lim_{k \to \infty} f_k(x) = f(x), a.e. x \in E.$$ If $\exists F(x)$, a Lebesgue integrable function defined on $E \subset \mathbb R^n$ such that $$|f_k(x)| \le F(x), a.e. x \in E \ (k=1,2,3,...),$$ then $$\lim_{k \to \infty} \int_{E} f_k(x)dx = \int_{E}f(x)dx.$$ Ususally, we call $F(x)$ is a dominated function.
Exercise2: Suppose $f \in L(E)$ and $f(x) > 0( \forall x \in E)$ then show $$\lim_{k \to \infty} \int_{E}(f(x))^{\frac{1}{k}}dx = m(E).$$
My trial is:Let $A = \{x \in E: f(x) < 1 \}$ then $$\lim_{k \to \infty} \int_{E}f^{\frac{1}{k}}(x)dx = \lim_{k \to \infty} \int_{A}f^{\frac{1}{k}}(x)dx + \lim_{k \to \infty} \int_{E-A}f^{\frac{1}{k}}(x)dx.$$ Then I got stuck here coz I did't know how to deal with limit in front of an integral.
Solution: Go on doing what I have done above, that is $$\lim_{k \to \infty} \int_{A}f^{\frac{1}{k}}(x)dx + \lim_{k \to \infty} \int_{E-A}f^{\frac{1}{k}}(x)dx = \int_{A} \lim_{k \to \infty} f^{\frac{1}{k}}(x)dx + \int_{E-A} \lim_{k \to \infty} f^{\frac{1}{k}}(x)dx = \int_{A} 1 dx + \int_{E-A}1dx = m(E).$$ So I think probably it has taken dominated convergence theorem into effect. However, what are the two dominated functions for $f(x)$ on $A$ and $E-A$, respectively? Coz I think the exercise doesn't say $m(E) < +\infty.$