1. ## Lebesgue Integral

Hi everyone
I coudnt have solved this problem, I hope you can give me some idea.

Let $\displaystyle E$ a measurable set and $\displaystyle f \in L_1(E)$, $\displaystyle f > 0$.
Prove that $\displaystyle \lim_{ n \rightarrow \infty} \int_E f^{1/n} d\mu = \mu(E)$.
Thanks
everk.

2. Notice that for all $\displaystyle x\in E$ we have $\displaystyle |f^{\frac{1}{n}}(x)|\leq 2|f(x)|$ and apply the most well known theorem for the Lebesgue integral.

3. Originally Posted by Jose27
Notice that for all $\displaystyle x\in E$ we have $\displaystyle |f^{\frac{1}{n}}(x)|\leq 2|f(x)|$ and apply the most well known theorem for the Lebesgue integral.
Sorry for this being so late, but this answer is wrong. The inequality is not valid on all of E. The correct approach is to consider two cases:

1- If E has finite measure apply the inequality above for the set on which f>1 and apply dominated convergence, and on the set on which $\displaystyle f\leq 1$ apply the monotone convergence theorem with bound m(E)

2- If E has infinite measure you have to prove that $\displaystyle \int_{E}f_n \rightarrow \infty$ for which just use that pointwise convergence implies convergence in measure so that you can bound below your seuqence (of integrals) by somethig which goes to infinity.