Given a function $f:\mathbb{R} \rightarrow \mathbb{R}$ derivable, is the function $$\sum_{n=1}^{\infty}\frac{f(x)^n}{n^n}$$ measurable?
Here my attempt:
Let $$S_n(x)=\sum_{k=1}^{n} \frac{f(x)^k}{k^k},$$ since it is the finite sum of measurable functions, $S_n(x)$ is measurable $\forall n \in \mathbb{N}$.
And $\forall x_0 \in \mathbb{R}$, $\exists \lim_{n\rightarrow\infty} S_n(x_0)=\sum_{n=1}^{\infty}\frac{f(x_0)^n}{n^n}$, because of the root test and the definition of function is convergent. So it's well-defined for every $x_0 \in \mathbb{R}$.
My professor have proved the theorem that establishes:
Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions and $D = \{x\in \mathbb{R}; \exists \lim_{n\rightarrow\infty} f_n(x)\}$, then the function $$f:D\rightarrow\mathbb{R}$$ defined as $f(x)=\lim_{n\rightarrow\infty}f_n(x)$ is measurable too.
With this theorem is easy to conclude the exercise.
It will be helpful any correction or another solution to the problem.
Thanks for everyone!