Let $f\in L^1_\text{loc}((0,\infty))$. If I prove that there exists $$ \lim_{\varepsilon\to0} \int_\varepsilon^T f(t)\, dt \,<\infty$$ for a given $T\in(0,\infty)$, can I conclude that $f\in L^1([0,T])$ ? Namely, $$ \int_0^T|f(t)|\,dt\,<\infty\ ?$$
I suspect there could be some problem related to the difference between Lebesgue and Riemann integrals.
No. The idea behind the following counterexample is that $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ is finite while $\sum_{n=1}^{\infty}\frac{1}{n}$ is not. So, all we have to do is make $f$ be a piece-wise constant function with appropriate values so that its integral over finite intervals equal the partial sums of these series.
More explicitly, let $f$ be the function whose restriction, for each integer $n\geq 1$, to the interval $\left(\frac{1}{n+1},\frac{1}{n}\right)$ is equal to $(-1)^n(n+1)$, and we set $f$ to $0$ outside the union of these intervals. Then, we have that $\int_{1/(n+1)}^{1/n}f(t)\,dt=\frac{(-1)^n}{n}$. Then, $f\in L^1_{\text{loc}}((0,\infty))$, because if we restrict to a compact subset, then $f$ is bounded, hence Lebesgue-integrable there. Also, by construction, \begin{align} \lim_{\epsilon\to 0^+}\int_{\epsilon}^1f(t)\,dt&=\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \end{align} but \begin{align} \int_0^{1}|f(t)|\,dt&=\sum_{n=1}^{\infty}\frac{1}{n}=\infty. \end{align}