If $\;f\;:[M,+\infty)\rightarrow \mathbb R\;$ is a non-negative $\;L_{loc}^1\;$function such that $\;\int_{M}^{\infty} f(s) \;ds =+\infty\;$ for $\;M \gt 0\;$, under which circumstances is it true to claim that $\;\int_{M}^{N} f(s) \;ds \lt +\infty\;$ for $\;N \gt M \gt 0\;$?
Although the above question seems intuitively always right, the following example came to my mind:
$\; \int_{0}^{\infty} \frac {1}{x} \;dx=+\infty=\int_{0}^{1} \frac {1}{x} \;dx\;$
Is this a counterexample for the initial question or $\;f(x)=\frac{1}{x}\;$ doesn't satisfy the above assumptions since $\;\frac{1}{x}\;$ is defined on $\; (0,+\infty) \;$ and not on $\; [0,+\infty) \;$?
Thanks in advance!!
The problem isn't that $f$ is not defined at $0$ - we could define it arbitrarily there - it is that $f\colon x \mapsto \frac{1}{x}$ isn't locally integrable at $0$, however we define it at $0$:
$$\int_0^{\varepsilon} \frac{1}{x}\,dx = +\infty$$
for all $\varepsilon > 0$.
The hypothesis is that we have a locally integrable function, and that implies that the function is integrable over all compact subsets of $[M,+\infty)$. Since $[M,N]$ is compact for all $M < N < +\infty$, the local integrability of $f$ guarantees
$$\int_M^N f(s)\,ds < +\infty$$
for all $M < N < +\infty$.