# Does $\lim_{\varepsilon\to0} \int_\varepsilon^T f(t)\, dt \,<\infty$ imply $\int_0^T|f(t)|\,dt\,<\infty$?

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.

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

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}
• thank you, nice example. I guess the same can be done with a suitable "sin-like" function, if one requires that $f$ is continuous Commented Jul 9, 2021 at 17:15
• @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. Commented Jul 9, 2021 at 17:21