let $X$ be of a finite measure & $f\in L^1$, $f(x) \neq 0$ a.e. Show if $E \subset X$: $ \lim_{n \to \infty} \int_E|f|^{1/n} d\mu = \mu(E)$

Question

I was trying to solve the next question:

let $X$ be a finite measure space ( $\mu(X)<\infty$) and let $f\in L^1(X, \mu)$, $f(x) \neq 0$ almost everywhere.
Show that for each measurable subset $E \subset X$:
$$ \lim_{n \to \infty} \int_E|f|^{\frac{1}{n}} d\mu = \mu(E)$$

My idea for a solution is to use Fatou's lemma:
In one direction:
$$ \int_E \liminf_{n \to \infty} |f|^\frac{1}{n} d\mu \leq \liminf_{n \to \infty} \int_E |f|^\frac{1}{n} d\mu $$ and
$$ \int_E \liminf_{n \to \infty} |f|^\frac{1}{n} = \int_E 1 d\mu = \mu(E)$$ So we get:
$$ \mu(E) \leq \liminf_{n \to \infty} \int_E |f|^\frac{1}{n} d\mu $$ In the other direction, I thought of maybe saying that we know there is an $\varepsilon > 0$
And an $N \in \mathbb{N}$ so for all $n > N$ we get that $1 + \varepsilon > |f|^\frac{1}{n}$ which means $1 + \varepsilon - |f|^\frac{1}{n} > 0$
and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to $\mu(E)$

Is it valid? Am I missing something?
If I do, what can I do to prove the other direction?
Thank you!

Answer 1

Lebesgue dominated convergence theorem will do:

Note that $|f|^{1/n}\leq|f|+1$ and $|f|+1\in L^{1}(\mu)$ since $\mu$ is of finite measure. Then interchange of integral with limit is fine.