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When I need to prove that $\int_{[0,\infty)}fd\mu<+\infty$ I usually think that if $\lim_{t\to\infty}\int_{[0,t]}fd\mu<\infty$ then $\int_{[0,\infty)}fd\mu$ must converge, but someone told me that argument was not correct (Cosider $\mu$ the Lebesgue measure).

That is why I am looking for an example of a continuous function $f:[0,\infty)\longrightarrow\mathbb R$ such that $\lim_{t\to\infty}\int_{[0,t]}fd\mu$ exists but $\int_{[0,\infty)}fd\mu$ is not defined.

So far I have not come up with any.

Harry Stone
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    You misunderstand what they mean. consider $f=1, \forall ,x\in \mathbb{R}$. then $\int_{[0,t]} 1 , d\mu = t< \infty \ \forall t>0$ but it does obviously not converge. – Nikolaj Pedersen Dec 21 '21 at 10:46
  • @NikolajPedersen Sorry my question was not well. Now is ok. Your examample don't verify $lim_{t\to\infty}\int_{[0,t]}fd\mu<\infty$ – Harry Stone Dec 21 '21 at 11:09

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Let $f(x)=\frac {\sin x} x$ if $x >0$ and $f(0)=1$. Then $\lim_{t \to \infty} \int_{[0,t]}f(x)dx=\frac {\pi} 2$. But $\int_{[0,\infty)} f(x)dx$ does not exist in Lebesgue sense. It is a general fact in Measure Theory that if $\int fd\mu$ is finite then $\in |f|d\mu <\infty$. In this example it is known that $\int |f(x)|dx=\infty$. Hence, it follows that neither $f^{+}$ nor $f^{-}$ is not integrable and $\int fd\mu$ is not defined.

Kavi Rama Murthy
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A classical example is the Dirichlet integral $\lim\limits_{t\to\infty}\int\limits_0^t\frac{\sin(x)}{x}\, dx = \pi/2$ although the function is not Lebesgue integrable. If you allow limits of integrals over $[-t,t]$, a trivial example is $\lim\limits_{t\to\infty}\int\limits_{-t}^t x dx =0$.

Jochen
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  • I did not notice that Kavi Rama Murthy has beaten me by some minutes. His answer should hus the accepted one. – Jochen Dec 21 '21 at 11:44