Lebesgue integral question concerning orders of limit and integration I've got a hand-in question in a pure analysis course that I was hoping I might get a hint on - having difficulty coming up with a decent approach. 
The question:
Let $(X,\Sigma,\mu)$ be a measure space and let $f:X\rightarrow [0,\infty]$ be a measurable function such that $$\int_X f(x)d\mu(x)=A,$$ for some $0<A<\infty$.
If $\alpha>0,$ show
$$\lim_{n\rightarrow\infty} \int_X n\log\left(1 + \left(\frac{f(x)}{n}\right)^{\alpha} \right)d\mu(x)=\begin{cases}
\infty&\mbox{if }0<\alpha<1\\\
A&\mbox{if } \alpha=1\\\
0&\mbox{if }\alpha>1.
\end{cases}$$
My attempt at a solution only comes as far as the first part:
\begin{align*}
g(x,n)&=n\log(1 + [f(x)/n]^{\alpha})\\
&=n\cdot \sum_{m=1}^{\infty}(-1)^{m+1}[f(x)/n]^{\alpha m}/m\\
&=\sum_{m=1}^{\infty}(-1)^{m+1}\cdot \frac{f(x)^{\alpha m}}{m\cdot n^{\alpha*m-1}} \\
&= \frac{f(x)^\alpha}{n^{\alpha-1}}+\sum_{m=2}^{\infty}(-1)^{m+1}\frac{f(x)^{\alpha m}}{m\cdot n^{\alpha m-1}},
\end{align*}
 which is increasing  in $n$ for $\alpha<1$ (this is a bit handwavy, but I can't seem to figure out how to show it in a strict manner). Thus, we can apply the Monotone Convergence Theorem to move the limit inside the integrand, transform $n=1/t$, use a bit of L'hopitals rule, and get that this limit is diverging for any $f(x)$ and $\alpha<1$ (and $f(x)$ for $\alpha=1$, zero for $\alpha>1$). But how do I go about proving that I can switch limit and integrand in these other cases, or is there any other simple way to prove it? Any hints would be much appreciated!
Many thanks in advance
 A: Let $I_n(\alpha):=\int_X n\log\left(1+\left(\frac{f(x)}n\right)^{\alpha}\right)d\mu (x)$. If $0<\alpha<1$, since $\int_X fd\mu>0$ we can find $\beta>0$ such that $\mu(\{f\geq \beta\})>0$. Therefore we have 
\begin{align*}
I_n(\alpha)&\geq\mu(f\geq \beta)n\log\left(1+\left(\frac{\beta}n\right)^{\alpha}\right) \\
&\geq \mu(f\geq \beta)n\left(\left(\frac{\beta}n\right)^{\alpha}-\frac 12\left(\frac{\beta^2}{n^2}\right)^{\alpha}\right)\\
&=\mu(f\geq \beta)n^{1-\alpha}\left(\beta^{\alpha}-\frac{\beta^{2\alpha}}{2n^{\alpha}}\right)
\end{align*}
and it converges to $+\infty$. 
If $\alpha=1$, the use of the inequality $x-\frac{x^2}2\leq \ln(1+x)\leq x$ allow us to apply the dominated convergence theorem.
If $\alpha>1$, use the measure $\nu=f\mu$ which is a probability measure. We have
$$I_n(\alpha)=\int_X\phi\left(\frac n{f(x)}\right)d\nu,$$
with $\phi(x)=x\ln (1+x^{-\alpha})$. Since $\phi'(x)=\ln(1+x^{-\alpha})-\alpha\frac{x^{-\alpha}}{1+x^{-\alpha}}$ and $\phi''(x)=\frac{-\alpha x^{-1-\alpha}}{1+x^{-\alpha}}-\frac 1{(1+x^{-\alpha})^2}(-\alpha x^{-\alpha-1})\alpha=-\alpha\frac{x^{-1-\alpha}}{1+x^{-\alpha}}x^{-\alpha}<0$ for $x>0$, $\phi$ is concave. Jensen's inequality yields
$$I_n(\alpha)=\int_X\phi\left(\frac n{f(x)}\right)d\nu\leq \phi\left(\int_X\frac n{f(x)}d\nu\right)\leq \left(\int_X\frac n{f(x)}d\nu\right)^{1-\alpha},$$
so it works if $(X,\Sigma,\mu)$ is finite. I guess we can solve the case $\sigma$-finite using the dominated convergence theorem for the counting measure.
