Lebesgue integral of power 1/n

Question

Let $f: X \rightarrow R$ be integrable.

WTS:

$$ \lim _{n \to \infty} \int |f|^{1/n} d{\mu}= \mu( x | f(x) \neq 0)$$

My attempt:

$$\lim_{n \to \infty} |f|^{1/n}= \chi_{\{x: f(x) \neq 0\}}.$$ By apply Fatou's lemma on $|f|^{1/n}$, we get $$ \int \liminf |f|^{1/n} = \mu( x | f(x) \neq 0) \leq \liminf \int|f|^{1/n}= \lim \int |f|^{1/n}.$$

I am not able to show the reverse inequality. I would like to use LDCT somehow,but $|f|^{1/n} \leq |f|$,only if $|f|\geq1$. In the other case $1 \geq|f|^{1/n} \geq|f|$ and since the measure space has no restriction, I cannot show that $|f|^{1/n}$ is integrable.

Can somebody please help?

Answer 1

Since $|f|^{1/n}\chi_{|f|\le 1}$ increases to $\chi_{0 < |f| \le 1}$, the monotone convergence theorem yields $\lim_{n\to \infty} \int |f|^{1/n} \chi_{|f| \le 1}\, d\mu = \int \chi_{0 < |f| \le 1} = \mu(\{x : 0 < |f(x)| \le 1\})$.

Now $|f|^{1/n}\chi_{|f| > 1}$ decreases to $\chi_{|f| > 1}$ and $\int |f|\chi_{|f| > 1}\, d\mu \le \int |f|\, d\mu < \infty$, so another application of the monotone convergence theorem produces $\lim_{n\to \infty} \int |f|^{1/n}\chi_{|f| > 1}\, d\mu = \int \chi_{|f| > 1}\, d\mu = \mu(\{x : |f(x)| > 1\})$.

The result is obtained by adding to the two limit equations.

Answer 2

Suppose $|f|>1$. Let $$ a_n=|f|^{\frac1n}-1. $$ Then $$ |f|=(a_n+1)^n\ge \frac{n(n-1)}{2}a_n^2 $$ and hence $$ 0\le a_n\le \sqrt{\frac{2|f|}{n(n-1)}}\le \sqrt{\frac{2}{n(n-1)}}|f|. $$ So $$ |f|^{\frac1n}\le1+a_n\le1+\sqrt{\frac{2}{n(n-1)}}|f|\le1+|f| \tag1$$ for $|f|>1$. For $|f|\le1$, one has $$ |f|^{\frac1n}\le1. \tag2$$ Let $g=(1+|f|)\chi_{f\neq0}$. Combining (1) and (2) gives $$ |f|^{\frac1n} \le g $$ on $X$. Note that $g$ is Lebesgue integrable. By Lebesgue's DCT, one has $$ \lim\int |f|^{1/n} d{\mu}=\lim\int_{f\neq0} |f|^{1/n} d{\mu}=\int_{f\neq0} \lim|f|^{1/n} d{\mu}=\mu( x | f(x) \neq 0). $$