When $\mu(X) < \infty$
$\{f_n\} \subseteq L^1(\mu)$, $f_n \to f$ uniformly, then $f \in L^1(\mu)$
is a true statement by If $\mu(X)<\infty$ then $f \in L^1(\mu)$ and $\int f_n \to \int f$ - An exercise question
What if $\mu(X) = \infty$?
My first thought was to let $X = (0, \infty)$ under the Lebesgue measure, define $f_n = (\frac{1}{x})^{1 + 1/n}$ and argue that while for all $n$, $\int |f_n| < \infty$, $\int |\frac{1}{x}| = \infty$.
However, it is not the case that $f_n \to f$ uniformly.
Can someone propose a counter-example where $X = \mathbb{R}$ and the Lebesgue measure is used?
Just take your favorite continuous non-$L^1$ function that vanishes at $\infty$, such as $\frac 1 x \chi_{\{x > 1\}}$, and cut it off on a large interval. You get a sequence such as
$$f_n(x) = \frac 1 x \chi_{\{1 < x < n\}}.$$