Let $(X, \mathcal{A}, \,u)$ be a measurable space and $f: X \rightarrow [0, +\infty]$ a measurable function such that $\int_Xf \mu =1$. Show that $$\lim_{n\to\infty} \int_X n\ln\left(1 + \frac{f}{n}\right)\, d\mu = 1$$
I am new to measure theory, so I still lack any intuition in it. I managed to bound this integral by 1. For a fixed $n \in \mathbb{N}^*$, we have: $$ \int_X nln(1 + \frac{f}{n})d\mu \leq \int_X n( \frac{f}{n})d\mu = \int_xfd\mu =1$$
Now, this is not sufficient, and I wonder two things: Can I say, that for an $n \in \mathbb{N}$ big enough, we have $|\frac{f}{n}|<1$ even though $f$ could take the value of $+\infty$? And secondly, if the answer to the first question is yes, could I then use Taylor series? This way I could appriximate $\ln(1 + \frac{f}{n})$ to $\frac{f}{n} + o(\frac{f^2}{n^2})$.
We have $\log(1+u)\leq u$ for $u\geq 0$, so $\log(1+f/n)\leq f/n$, so $n\log(1+f/n)\leq f$. Now $n\log(1+f/n)\rightarrow f$ a.e. as $n\rightarrow\infty$, so by Lebesgue Dominated Convergence Theorem we have $\lim_{n\rightarrow\infty}\displaystyle\int_{X}n\log(1+f/n)d\mu=\displaystyle\int_{X}fd\mu=1$.
Note that $\displaystyle\int_{X}fd\mu=1<\infty$, then $f<\infty$ a.e.