Calculate $\lim_{n\rightarrow +\infty}n\int_X\log\biggl(1+\frac{f(x)}{n}\biggl)d\mu$ Let $\mu$ be a strictly positive measure, and let $f:X\rightarrow[0,\infty]$ be a an function such that $\int_Xfd\mu=1$. Calculate the following limit:
$\lim_{n\rightarrow \infty}n\int_X\log\biggl(1+\frac{f(x)}{n}\biggl)d\mu$
 A: Note that $\left|n\log\left(1+\frac{f(x)}n\right)\right|\le|f(x)|$, so we can use $|f(x)|$ as a dominating function for Dominated Convergence.
Pointwise,
$$
\lim_{n\to\infty}n\log\left(1+\frac{f(x)}n\right)=f(x)
$$
Therefore, by Dominated Convergence, we have
$$
\lim_{n\to\infty}n\int_X\log\left(1+\frac{f(x)}n\right)\,\mathrm{d}\mu
=\int_Xf(x)\,\mathrm{d}\mu=1
$$
A: My approach:
Since $\log(x+1)\leq x$ for all $x \in [0,\infty]$ and $f \leq 0$ then for every  $n>0$: $n\log\biggl(1+\frac{f(x)}{n}\biggl) \geq 0$ and therefore: $\biggl |n\log\biggl(1+\frac{f(x)}{n}\biggl)\biggl| \leq n \cdot \frac{f(x)}{n}=f(x)$.
Morevoer, because $f$ is non-nogative, then $\int_X|f|d\mu=\int_Xfd\mu=1<\infty$.
$\implies f \in L^1(X)$. i.e, integrable over X.
Clearly, from the monotony of the integral:
$\int_X\biggl|n\log\biggl(1+\frac{f(x)}{n}\biggl)\biggl|d\mu \leq \int_Xf(x)d\mu=1<\infty.$ So, $n\log\biggl(1+\frac{f(x)}{n}\biggl) \in L^1(X)$ as well.
Now, we notice that: $\lim_{n\rightarrow\infty}n\log\biggl(1+\frac{f(x)}{n}\biggl)=\lim_{n\rightarrow\infty}\log\biggl(\biggl(1+\frac{f(x)}{n}\biggl)^n\biggl) = \log\biggl(\lim_{n\rightarrow\infty}\biggl(1+\frac{f(x)}{n}\biggl)^n\biggl)$ 
$= \log\biggl(\exp(f(x))\biggl)=f(x)$
$\implies$ By the dominant convergence theorem, we get:
$\lim_{n\rightarrow \infty}n\int_X\log\biggl(1+\frac{f(x)}{n}\biggl)d\mu = \lim_{n\rightarrow \infty}\int_Xn\log\biggl(1+\frac{f(x)}{n}\biggl)d\mu$ 
$= \int_X\lim_{n\rightarrow \infty}n\log\biggl(1+\frac{f(x)}{n}\biggl)d\mu = \int_Xf(x)d\mu = 1$
Is my solution is fine? Did I miss anything that is worth explaining?
I would love for some feedback!
Thank you very much :)
