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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.

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  • $\begingroup$ The essential point is that $\displaystyle \int_\varepsilon^T f(t)\, dt$ may converge CONDITIONALLY as $\varepsilon\downarrow0. \qquad$ $\endgroup$ Commented Jul 9, 2021 at 18:24

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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}

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  • $\begingroup$ thank you, nice example. I guess the same can be done with a suitable "sin-like" function, if one requires that $f$ is continuous $\endgroup$
    – tituf
    Commented Jul 9, 2021 at 17:15
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    $\begingroup$ @tituf yes in fact, $\lim\limits_{R\to\infty}\int_0^{R}\frac{\sin u}{u}\,du$ is finite but $\int_0^{\infty}\left|\frac{\sin u}{u}\right|\,du$ is not, and so by making the substitution $u=\frac{1}{t}$, we see that $\lim_{\epsilon\to 0^+}\int_{\epsilon}^{\infty}\frac{1}{t}\sin\left(\frac{1}{t}\right)\,dt$ is finite while $\int_0^{\infty}\left|\frac{1}{t}\sin\left(\frac{1}{t}\right)\right|\,dt$ is not. So, $f(t)=\frac{1}{t}\sin\left(\frac{1}{t}\right)$ works (define it however you wish at $t=0$). I didn't want to assume familiarity with this integral, hence I presented the above answer. $\endgroup$
    – peek-a-boo
    Commented Jul 9, 2021 at 17:21

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