im trying to prove the following:
Let $f$ be a non negative integrable function with $0<\int fd\mu=c<\infty$ and $0<\alpha<\infty$.
Prove that
$$\lim_{n\to\infty}\int n\log\left[1+\left(\frac{f}{n}\right)^\alpha\right]d\mu=\left\{\begin{array}{ll}\infty,&\text{ if }0<\alpha<1,\\c,&\text{ if }\alpha=1,\\0,&\text{ if }1<\alpha<\infty.\end{array}\right.$$
This is what i have:
Suppose $\alpha=1$, we take $f_n$ to be $$f_n=n\log\left(1+\frac{f}{n}\right)$$ then take the limit $$\lim_{n\to\infty} f_n=\lim_{n\to\infty} \log\left(1+\frac{f}{n}\right)^n=\log \lim_{n\to\infty}\left(1+\frac{f}{n}\right)^n=\log e^f=f.$$ So, if $f$ is the limit of $f_n$, so $$f_n\leq f, \forall n$$ Then by the dominated convergence theorem: $$\lim_{n\to\infty}\int f_nd\mu=\int\lim_{n\to\infty}f_nd\mu=\int fd\mu=c.$$
But i can't see how to approach the case when $0<\alpha<1$, and $1<\alpha$. So any help would be appreciated.
Your answer seems all right, you just need to be careful that from what you wrote it does not follow that $f_n\leq f$ you also need to know that the sequence is monotone increasing. It is also useful to assume that $f$ is bounded outside a measure zero set. I will help you with the limit part and maybe you can fill in the rest.
For the limit you want to look at the following:
$$e^{n\log(1+\frac{f^\alpha}{n^\alpha})}=(1+\frac{f^\alpha}{n^\alpha})^n$$
do the limit for $n\rightarrow\infty$ of both sides and use Hopital's rule with variable $n$. Then the limit of the LHS becomes:
$$\exp\left(\frac{\frac{1}{1+\frac{f^\alpha}{n^\alpha}}\times (-\frac{f^\alpha}{n^{\alpha+1}})}{-\frac{1}{n^2}}\right)=\exp\left(\frac{1}{1+\frac{f^\alpha}{n^\alpha}}\times \frac{n^2}{n^{\alpha+1}}\right)$$
now if $\alpha>1$ this limit goes to $1$ and if $\alpha<1$ it goes to infinity and you are taking the $\log$ of this limit.