Let $S$ be a bounded measurable subset of $\mathbb R$. Let $f \colon S → (0,\infty)$ be Lebesgue integrable. Prove that
$$\lim_{n\to\infty}\int_S\ f^{1/n} \;\mathrm{d}m = m(S)$$
Where $m(S)$ is the Lebesgue measure of $S$.
There is a hint that let $S = A \cup B$, where $A = f^{−1}((0, 1))$ and $B = f^{−1}([1,+\infty))$.
And the sequence of functions $h_n\ (n = 1, 2, 3, \dots),$ where $h_n(x) = ({f(x)})^{1/n}$.
I notice that $h_1(x)\leq h_2(x)\leq h_3(x) \leq \dots$
How can I apply monotone convergence theorem to the given hint.
If you let $a \in A$ then you notice that $(f(a))^{1/n}$ is a monotonically increasing sequence as $n \to \infty$. If $b \in B$ then $(f(b))^{1/n}$ is monotonically decreasing as $n \to \infty$, which you can modify to comply with the monotone convergence theorem. After you know that you look at $$ \int_S f^{1/n} \, dm = \int_A f^{1/n} \, dm + \int_B f^{1/n} \, dm $$ and take the limit.