# show that $f_k$ is lebesgue integratable

Let $$f : \mathbb{R}^n \to [0;\infty[$$ be a non-negative, lebesgue integratable function and for every $$k \in \mathbb{N}$$ let the functions $$f : \mathbb{R} \to \mathbb{R}$$ be $$f_k(x) = k\cdot \log(1+\frac{f(x)}{k})$$. Show that f is lebesgue integratable for every k and that $$lim_{k \to \infty} \int f_k dv_n = \int f dv_n$$.

So, my idea was to show that the set $$A_c := \{x \in \mathbb{R}: f_k(x) \geq c_k := k \cdot \log(1)\}$$ is measurable and then concluding f is measurable. But it seems hard to show that $$A_c$$ is either closed or open. I haven't thought about the second part yet, where one needs to show that the $$\lim$$ of $$f_k$$ for $$k \to \infty$$ is equal to $$f$$. Help there would be appreciated too since I have no idea how to solve that either.

• $v_n$ is the notation for lebesgue integration. I fixed a typo if that is what you were wondering about. – Max Dec 9 '19 at 7:46

$$k \log (1+\frac a k) \to a$$ for every real number $$a$$.
From the inequality $$\log (1+x) \leq x$$ for $$x \geq 0$$ we see that $$0\leq f_k(x) \leq f(x)$$. Also $$f_k (x) \to f(x)$$ for all $$x$$. Hence DCT can be applied.
• why exactly is $f_k$ now measurable? Does $0 \leq f_k(x) \leq f(x)$ implie the set $A_c$ is closed? – Max Dec 9 '19 at 11:21
• @Max $f_n$ is of the form $g(f(x))$ where $g$ is a continuous function. This implies that $f_k$ is measurable. – Kavi Rama Murthy Dec 9 '19 at 11:52
HINT: for any $$c\geqslant 0$$ the sequence $$(c_n)$$ defined by $$c_n:=\left(1+\frac{c}{n}\right)^n$$ is increasing and converge to $$e^c$$.