This may be simple, but I want to know if my reasoning is ok. I came across a problem whose essential set up is:
let $f_k$ be a sequence of functions in $L^1(\mathbb{R})$ (Lebesgue integrable functions on $\mathbb{R}$). Suppose that
$$\displaystyle \lim_{t \to \infty} f_k(t) = 0 \text{ for all } k \in \mathbb{N} \quad \text{ and } \quad \displaystyle \sum\limits_{k=0}^\infty f_k \in L^1(\mathbb{R})$$
We of course have that $\displaystyle \sum\limits_{k=0}^\infty \lim_{t \to \infty} f_k(t) = 0$. Where I'm having a bit of doubt, is proving that
$$\lim_{t \to \infty} \sum\limits_{k=0}^\infty f_k(t) = 0, \tag{1}$$
my argument is: since for any $n \in \mathbb{N}$, we have that $\displaystyle\lim_{t \to \infty} \sum\limits_{k=0}^n f_k(t) = \sum\limits_{k=0}^n \lim_{t \to \infty} f_k(t) = 0,$ so $(1)$ follows trivially from this last observation.
Do you see something wrong in my argument?
In the best case, I'm asking an obvious question. In the worst case I'm missing something very badly...
PS. I know I did not used much the fact that $f_k \in L^1(\mathbb{R})$ and the sum converges in $L^1$, but I point it out in case I'm missing something.
EDIT. I found this post: Conditions for taking a limit into an infinite sum asking a similar question, but the OP mentions that uniform convergence is needed.
The chosen answer says you have to use the DCT, maybe that's the argument I need, I'll check it out, meanwhile I would appreciate any feedback.
The limit in (1) need not exist.
Let $f_k = \chi_{[k, k + 1 / k^2]}$ be the indicator function of the interval $[k, k + 1 / k^2]$ ($k \geq 1$). These are obviously Lebesgue-integrable, $\lim_{t \to \infty} f_k(t) = 0$ and $\int_{\mathbb{R}} \sum_{k = 1}^\infty f_k \mathrm{d}\mu = \frac{\pi^2}{6}$ (since the integral computes the Basel problem sum $\sum_{k = 1}^{\infty} \frac{1}{k^2}$), whence $f = \sum_{k = 1}^\infty f_k \in L^1(\mathbb{R})$. But for every $r > 0$, there exists points $x_0, x_1 > r$ (for instance, $x_1 = \lceil r \rceil$ and $x_0 = \lceil r \rceil + 11 / 4$) such that $f(x_0) = 0$ and $f(x_1) = 1$, whence $\lim_{t \to \infty} f(t)$ fails to converge.