# I was trying to solve the next question: let X be a finite measure space ( &#x03BC;<!--

I was trying to solve the next question:
let $X$ be a finite measure space ( $\mu \left(X\right)<\mathrm{\infty }$ ) and let $f\in {L}^{1}\left(X,\mu \right)$, $f\left(x\right)\ne 0$ almost everywhere.
Show that for each measurable subset $E\subset X$:
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{E}|f{|}^{\frac{1}{n}}d\mu =\mu \left(E\right)$
My idea for a solution is to use Fatou's lemma:
In one direction:
${\int }_{E}\underset{n\to \mathrm{\infty }}{lim inf}|f{|}^{\frac{1}{n}}d\mu \le \underset{n\to \mathrm{\infty }}{lim inf}{\int }_{E}|f{|}^{\frac{1}{n}}d\mu$
and
${\int }_{E}\underset{n\to \mathrm{\infty }}{lim inf}|f{|}^{\frac{1}{n}}={\int }_{E}1d\mu =\mu \left(E\right)$
So we get:
$\mu \left(E\right)\le \underset{n\to \mathrm{\infty }}{lim inf}{\int }_{E}|f{|}^{\frac{1}{n}}d\mu$
In the other direction, I thought of maybe saying that we know there is an $\epsilon >0$
And an $N\in \mathbb{N}$ so for all $n>N$ we get that $1+\epsilon >|f{|}^{\frac{1}{n}}$ which means $1+\epsilon -|f{|}^{\frac{1}{n}}>0$
and use Fatou's lemma again on the expression above to get the lim sup smaller or equal to $\mu \left(E\right)$

Is it valid? Am I missing something?
If I do, what can I do to prove the other direction?
Thank you!
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Mathias Patrick
Lebesgue dominated convergence theorem will do:
Note that $|f{|}^{1/n}\le |f|+1$ and $|f|+1\in {L}^{1}\left(\mu \right)$ since $\mu$ is of finite measure. Then interchange of integral with limit is fine.