I am currently working on the following task
Let $f \in \mathcal{M_+}(\mathbb{R},\overline{\mathbb{R}})$ an $\mu$ is a borel-measure on $\mathbb{R}.$ Show that,
$$\lim \limits_{n \to \infty} n \int_{-\infty}^\infty log{\left(1 + \frac{1}nf\right)} \ \ \ d\mu \ = \ \int_{-\infty}^\infty f \ d\mu. $$
We defined $\mathcal{M_+}(\mathbb{R},\overline{\mathbb{R}})$ as the set of measurable maps $f:\mathbb{R} \to [0,\infty].$
because $f \in \mathcal{M_+}(\mathbb{R},\overline{\mathbb{R}})$ there exist a sequence of functions $({f_n})_n \subset \mathcal{M_o}(\mathbb{R},\mathbb{R})$ with $0\le f_n \uparrow f$
- we defined $\mathcal{M_o}(\mathbb{R},\mathbb{R})$ as the set of step functions $f:\mathbb{R} \to \mathbb{R}$. Step functions are borel-measurable and have the shape of $\sum_{i=1}^{m} c_i \mathcal{X}A_i$
How do I start with the task? I would appreciate a hint or a little help.
What bothers me is the upper and lower limits of the integral. Do I need to change the upper and lower limits? If yes, how I can do that?
To the right side of the equation I am not sure if I should construct a certain f to $\lim \limits_{n \to \infty} n \int_{-\infty}^\infty log{\left(1 + \frac{1}nf\right)}$.
Can I replace the f on the left side of the equation with a sequence of functions $(f_n)_n$. If so, what is the advantage of doing this?
kind regards,
WomBud