1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $v_n$ is the notation for lebesgue integration. I fixed a typo if that is what you were wondering about. $\endgroup$ – Max Dec 9 '19 at 7:46
1
$\begingroup$

$k \log (1+\frac a k) \to a$ 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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ why exactly is $f_k$ now measurable? Does $0 \leq f_k(x) \leq f(x)$ implie the set $A_c$ is closed? $\endgroup$ – Max Dec 9 '19 at 11:21
  • $\begingroup$ @Max $f_n$ is of the form $g(f(x))$ where $g$ is a continuous function. This implies that $f_k$ is measurable. $\endgroup$ – Kavi Rama Murthy Dec 9 '19 at 11:52
1
$\begingroup$

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 $e^c$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.