# Is the function $\sum_{n=1}^{\infty}\frac{f(x)^n}{n^n}$ measurable?

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!

• Your argument for the pointwise convergence of $(S_n)$ is incorrect. That has nothing to do with measurability/continuity/differentiability of $f$. Once you give a correct argument for the pointwise convergence, you have a correct proof. – Daniel Fischer Jan 22 '18 at 12:35
• I skipped the proof of the convergence of the series for every $x_0$ since it's trivial using the root test for numerical series. Are you referring to this? @DanielFischer – DrinkingDonuts Jan 22 '18 at 12:39
• Yes. You probably don't need to provide all details, but you ought to mention the reason why the series converges everywhere. And the well-definedness at all $x_0 \in \mathbb{R}$ has nothing to do with differentiability, that's part of what being a function is. – Daniel Fischer Jan 22 '18 at 12:44
• Many thanks for the details you provided! @DanielFischer – DrinkingDonuts Jan 22 '18 at 12:55