$\lim_{n}n\int_{\Omega}\ln\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)d\mu$ I found a follow exercise:
Let $f:\Omega\to\mathbb{R}$ a positive function such that $0<\displaystyle\int_{\Omega}f\,d\mu<\infty$. If $\alpha>0$ is a constant, show that
$$
\lim_{n}n\int_{\Omega}\ln\left(1+\left(\frac{f}{n}\right)^{\alpha}\right)d\mu=\left\{\begin{array}{lcl} \infty&if&0<\alpha<1\\ \displaystyle\int_{\Omega}fd\mu&if&\alpha=1\\ 0&if&\alpha>1 \end{array} \right.
$$

I have achieved the following, if $\alpha=1$, then $f_n=\left(1+\frac{f}{n}\right)^n\nearrow f$, thus by monotonic convergence theorem $\lim_n \int_{\Omega}f_n\,d\mu=\int_{\Omega}f\,d\mu$. If $\alpha>1$, then $f_n\to 0$, but as for example to use the Convergence Theorem I dominate a integrable function $g$ such that $|f_n|\leq g$ for all $n$, but I can't see who this function would be. The same happens to me in case $0<\alpha<1$, because we have that $f_n\to\infty$, but in this case I don't know how to conclude that $\int_{\Omega}f\,d\mu=+\infty$ or perhaps this case that I am wondering about is totally immediate, I don't see it.

 A: If you want to argue with convergence theorems, you go as the following

When $0<\alpha<1$, $f_n$ is still increasing in $n$ hence you can use the Monotone Convergence Theorem to imply the limit.
When $\alpha>1$, note that because $$\lim_{x \rightarrow 0^+} \frac{\ln(1+x^{\alpha})}{x} =\lim_{x \rightarrow \infty} \frac{\ln(1+x^{\alpha})}{x}=0 $$
then there is a constant $C>0$ such that $$ \ln(1+x^{\alpha})\le Cx $$ for all $x$ positive, from which you can use DCT.

If you want to try something different, you can use the following identity.

Identity:
$$n \int \ln\left( 1+\left[ \frac{f}{n}\right]^{\alpha}\right)d\mu=\int_{0}^{\infty} \frac{\alpha \left( \frac{u}{n}\right)^{\alpha-1}}{1+\left( \frac{u}{n}\right)^{\alpha}} \mu( \{f>u\})du$$
and
$$ \int f \mu = \int_{0}^{\infty} \mu( \{ f>u\})du$$

A: I was confused originally on the use os square brackets in your OP.

*

*If $[\;]$ is used for the ingteger part functions, then the statement of the problem is not entirely true for $0<\alpha<1$, for example, if $f\in L^+_1(\mu)\cap L_\infty(\mu)$, then $\log\Big(1+\Big\lfloor \frac{f}{n}\Big\rfloor^\alpha\Big)=0$ for all $n> \|f\|_\infty$. In this case.

*I presume the problem is meant to say $\log\Big(1+\big(\frac{f}{n}\big)^\alpha\Big)$.


On the understating that $[\;]$ is meant as a parenthesis
If $0<\alpha<1$ then $n \log\Big(1+\big(\frac{f}{n}\big)^\alpha\Big)\geq \log(1+ n^{1-\alpha}f^\alpha)$ by the binomial theorem. As $\phi_n=\log(1+n^{1-\alpha}f^\alpha)$ is an increasing sequence of nonnegative measurable functions, then by monotone convergence
$$\lim_{n\rightarrow\infty}\int\phi_n\,d\mu=\int\lim_n\phi_n\,d\mu=\infty$$
If $\alpha\geq1$, then as
$$\lim_{x\rightarrow0}\frac{\log(1+t^\alpha)}{t}=\lim_{t\rightarrow\infty}\frac{\log(1+t^\alpha)}{t}=0$$
there exists a constant $C_\alpha>0$ such that $\log(1+t)\leq C_\alpha t$. Then
$$n\log\Big(1+\Big(\frac{f}{n}\Big)^\alpha\Big)\leq  C_\alpha f$$
Since $\lim_{n\rightarrow\infty}n\log\Big(1+\Big(\frac{f}{n}\Big)^\alpha\Big)=\lim_n\frac{1}{n^{\alpha-1}} f^\alpha$ which is $f$ if $\alpha=1$ and $0$ for $\alpha>1$, the conclusion follows by dominated convergence.
