I have the following problem:
Let $f:\Omega \rightarrow [0,\infty)$ be an integrable function. Show that for the functions $$g_n(x)=\log\left(1+\frac{f(x)}{n}\right)$$ we have $$\lim_{n\rightarrow \infty} n\int_\Omega g_n \,d\mu=\int_\Omega f\, d\mu$$
I have just shown that $g_n$ is integrable. Now I wanted to proof the above statement as follows:
Proof Let us first remark that $$\lim_{n\rightarrow \infty} n\int_\Omega g_n \,d\mu=\lim_{n\rightarrow \infty} \int_\Omega \log\left(\left(1+\frac{f(x)}{n}\right)^n\right) d\mu$$
Let us define now $h_n(x)=\log\left(\left(1+\frac{f(x)}{n}\right)^n\right)$, then $\lim_{n\rightarrow \infty} h_n(x)=f(x)$. Furthermore we can also see that $|h_n(x)|=h_n(x)\leq f(x)$ for all $n\geq 1$.
Since $f(x)$ is integrable by assumption we can use the dominated convergence theorem and get: $$\lim_{n\rightarrow \infty} \int_\Omega \log\left(\left(1+\frac{f(x)}{n}\right)^n\right)=\int_\Omega \lim_{n\rightarrow \infty}\log\left(\left(1+\frac{f(x)}{n}\right)^n\right)\stackrel{\text{$\log$ is continuous}}{=}$$ $$= \int_\Omega \log\left(\lim_{n\rightarrow \infty}\left(1+\frac{f(x)}{n}\right)^n\right)=\int_\Omega f(x)\, d\mu$$
Does this work so?
Thanks for your help.