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Let $(X,\mathcal{A}, \mu)$ be of finite measure and $f$ integrable on the measure space. Show that:

$\lim_{n \to \infty} \int_{X}|f|^{\frac{1}{n}}d\mu=\mu(\{f\neq0\})$

My ideas:

I initially thought of using the DCT but it is not necessarily true that $|f|^{\frac{1}{n}}\leq|f|$ so that goes out the window. Perhaps assuming $f$ is a simple function.

So we get: $\int_{X}f^{\frac{1}{n}}d\mu=\sum_{i=1}^{m}\alpha_{i}^{\frac{1}{n}}\mu(A_{i})$ and then $\lim_{n \to \infty}\sum_{i=1}^{m}\alpha_{i}^{\frac{1}{n}}\mu(A_{i})=\sum_{i=1}^{m}\lim_{n \to \infty}\alpha_{i}^{\frac{1}{n}}\mu(A_{i})=1\times\mu(\{f\neq0\})+0\times \mu(\{f =0\})=\mu(\{f\neq0\})$

It does not seem correct, as I have not use the finite measure of $\mu$ on $\mathcal{A}$

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Try using $1+|f|$ as a majorant instead of $|f|$.

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Hint:

Use the dominated convergence theorem.

If you can make the $\lim$ inside the integral you can use the fact that $\lim_{n\rightarrow\infty} \sqrt[n]{|x|} = 1$ whenever $x\not = 0$.

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