Proof that $\lim_{n\to\infty}\int_S\ f^{1/n} \, dm = m(S)$

Question

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.

Answer 1

You can apply monotone convergence to show $\int_A f^{1/n}\to m(A)$, and then dominated convergence to show $\int_B f^{1/n}\to m(B)$.