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. 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