I am trying to prove the following:
Let $(X,\Sigma, \mu)$ be a measure space and let $f:X\rightarrow [0,\infty)$ be an integrable function. Then for any real number $\alpha>1$, $$\int_X n \log\left(1+\left(\dfrac{f(x)}{n}\right)^\alpha\right)\:d\mu(x)\rightarrow 0$$ as $n\rightarrow \infty$.
Let's call function inside the integrand $f_n(x)$. I have already shown that the pointwise limit of $f_n$ is $0$ (this can be done using L'Hopital). Naturally, my plan is to use dominated convergence theorem and say that $$\lim_{n\rightarrow \infty} \int_X f_n\:d\mu= \int_X \lim_{n\rightarrow \infty} f_n\:d\mu=0.$$ But to do so, I need to show $f_n<g$ for some integrable function $g$, and I have yet to find such a $g$. I have noticed that since $\log (x+1)\le x$ for non-negative $x$, we have $$ f_n (x) \le n\left(\dfrac{f(x)}{n}\right)^\alpha=f(x)\left(\dfrac{f(x)}{n}\right)^{\alpha-1}.$$ I am not sure if this is a good move, but I tried integrating the thing on the right like this $$\int_X f(x)\left(\dfrac{f(x)}{n}\right)^{\alpha-1}\: d\mu(x)=\int_{\{f> n\}} f(x)\left(\dfrac{f(x)}{n}\right)^{\alpha-1}\: d\mu(x)+\int_{\{f\le n\}} f(x)\left(\dfrac{f(x)}{n}\right)^{\alpha-1}\: d\mu(x)$$ $$\le\int_{\{f> n\}} f(x)\left(\dfrac{f(x)}{n}\right)^{\alpha-1}\: d\mu(x)+\int_{\{f\le n\}} f(x)\: d\mu(x).$$ This looked promising at the time, the integral for $\{f\le n\}$ is finite by assumption, but I can't seem to kill off the integral for $\{f>n\}$. It seems to be intuitively clear that it should converge to $0$ if I let $n\rightarrow\infty$ but I'm not sure.
Any help will be apprieciated!
Define $$f_n(x):=n\log\left(1+\left(\dfrac{f(x)}{n}\right)^\alpha\right)\mathbf 1\{f\leqslant n\} \mbox{ and } $$ $$g_n(x):=n\log\left(1+\left(\dfrac{f(x)}{n}\right)^\alpha\right)\mathbf 1\{f\gt n\} .$$ Then for each $n$, $$0\leqslant f_n(x)\leqslant n\cdot (f(x)/n)^{a-1}f(x)/n\mathbf 1\{f\leqslant n\} \leqslant f(x).$$
In order to bound $g_n$ independently of $n$, we can use the fact that there exists a constant $C$ such that for each $t\geqslant 1$, $\log (1+t)\leqslant Ct^{1/a}$ (this is due to the fact that $\log(1+t) /t^{1/a}$ goes to $0$ as $t$ goes to infinity and $t\mapsto\log(1+t) /t^{1/a}$ is continuous on $[1,\infty)$.