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

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 $$\mathcal{A}$$

Try using $$1+|f|$$ as a majorant instead of $$|f|$$.
If you can make the $$\lim$$ inside the integral you can use the fact that $$\lim_{n\rightarrow\infty} \sqrt[n]{|x|} = 1$$ whenever $$x\not = 0$$.