Almost everywhere convergence and Lebesgue integrability of $\displaystyle\sum_{n = 1}^{+\infty} f_n (x)$

Question

I have the following exercise:

For $n \in \mathbb N$ consider the function $$f_n : \mathbb R \to \mathbb R, \quad f_n (x) := \begin{cases} \frac{\lvert\ln (n^7x^2)\rvert}{3+n^4x^2} & \text{if } x \ne 0 \\ 1 & \text{if } x=0\end{cases}$$ Then:

1. Discuss the integrability à la Lebesgue of the functions $f_n$.
2. Say if the series $\displaystyle\sum_{n=1}^{+\infty} f_n(x)$ converges amost everywhere in $\mathbb R$. In which case, if $C$ indicates the set of $x \in \mathbb R$ for which the series converges and we introduce the function $$s : C \to \mathbb R, \quad s(x) := \sum_{n=1}^{+\infty} f_n(x) ,$$ is $s$ a measurable function?
3. Does the equality $$\int_C s(x) \mathrm d x = \sum_{n=1}^{+\infty} \int_C f_n(x) \mathrm d x$$ hold? Is $s \in L^1 \mathbb R$ too?

The first point is addressed in a separate question in MSE, namely this one. Now, $\displaystyle\sum_{n=1}^{+\infty} f_n(0) = +\infty$, hence $C \subseteq \mathbb R \setminus \{0\}$. If we restrict to $x \ne 0$, we have $\displaystyle\sum_{n=1}^{+\infty} f_n(x)$ has the same behaviour of $\displaystyle\sum_{n=1}^{+\infty} \frac{\ln n}{n^4}$ which does converge. Consequently, the series of the exercise converges for every $x \in \mathbb R, x \ne 0$, and $C =\mathbb R \setminus \{0\}$. The series $s : C \to \mathbb R$ is measurable, since it the pointwise limit of a sequence of masurable functions.

The equality $\displaystyle\int_C s(x) \mathrm d x = \sum_{n=1}^{+\infty} \int_C f_n(x) \mathrm d x$ holds because of the of the Beppo-Levi Theorem. If $\lvert x \rvert \ge 1$, then $$\sum_{n=1}^{+\infty} f_n(x) \le \frac{7}{x^2} \underbrace{\sum_{n=1}^{+\infty} \frac{\ln n}{n^4}}_{< +\infty} + 2 \frac{\ln \lvert x \rvert}{x^2} \underbrace{\sum_{n=1}^{+\infty}\frac{1}{n^4}}_{< + \infty}$$ Here, right-hand side is integrable on $(-\infty, -1] \cup [1, +\infty)$. What about the integrability of $s$ on $[-1, 1]$?

By the way, is what I've done so far correct?

Answer 1

Yes, your work so far is correct. To address the integrability of s on [−1,1], we need to consider the behavior of the function near x = 0. To be integrable, the function must be bounded and have finite total variation on every finite interval. As we know that ln(n7x2) → −∞ as x → 0, fn(x) becomes arbitrarily large as x approaches 0, which indicates that the function is not integrable in a neighborhood of x = 0.

Therefore, s is not integrable on R, as it is not integrable on [−1,1].

Also, as s is not integrable, it does not belong to L1(R).