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I want to know if the following statement is true.

Assume that $g:\mathbb{R}\rightarrow\mathbb{R}$ is function such that any integral of the form $\int_{-\infty}^{t}g(x)dx$ is finite and there exists a limit $\lim_{t\to\infty}\int_{-\infty}^{t}g(x)dx$ that is finite.

Then $\int_{\mathbb{R}}g(x)dx$ exist and is finite.

How to prove it or find some counter-example?

Community
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Lukaszmat
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1 Answers1

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The claim is false.

I give a counterexample. I denote $d\mu(x)$ for Lebesgue integral and $dx$ for Riemann integral.

Take $$g(x)=\mathbf{1}_{x> 0}\frac{\sin(x)}{x}+\mathbf{1}_{x=0}$$ Note that there is no need to integrate on $(-\infty,0]$. Since this function is Riemann integrable on $[0,t]$ for all $t>0$ we have: \begin{align} \int_{[0,t]}\frac{\sin(x)}{x}\,d\mu(x) = \int^t_0 \frac{\sin(x)}{x}\,dx \end{align} Hence: \begin{align} \lim_{t\to\infty}\int_{[0,t]}\frac{\sin(x)}{x}\,d\mu(x)=\lim_{t\to\infty}\int^t_0 \frac{\sin(x)}{x}\,dx = \frac{\pi}{2} \end{align} Finite and all. However $\int_{[0,\infty)}\frac{\sin(x)}{x}\,d\mu(x)$ does not even exist, since $\frac{\sin(x)}{x}$ is not Lebesgue integrable on $[0,\infty)$. See this post for a proof.

Shashi
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  • I understand that $\int_{[0,\infty)}\left|\frac{sin(x)}{x}\right|d\mu(x)=\infty$, but it means that $\int_{[0,\infty)}\frac{sin(x)}{x}d\mu(x)$ does not exist? – Lukaszmat Dec 31 '17 at 15:44
  • @Lukaszmat yes, because what is the definition of lebesgue integrability? That $\int |f|, d\mu<\infty$. Remember? – Shashi Dec 31 '17 at 15:46
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    Yes, but maybe this integral exists and is equal $\infty$ or $-\infty$ – Lukaszmat Dec 31 '17 at 15:48
  • @Lukaszmat no it just does not exist. Lebesgue does not deal with not absolutely convergent integrals. Are you sure you know the definition of Lebesgue integral? – Shashi Dec 31 '17 at 15:50
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    For example. Assume that we have function $f(x)=1$ on real line. Then $\int_{\mathbb{R}}\left|f(x)\right|d\mu(x)=\infty$ but $\int_{\mathbb{R}}f(x)d\mu(x)$ exist and is equal $\infty$. For me exist does not mean that it is finite. – Lukaszmat Dec 31 '17 at 15:56
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    For me does not exist means that $\int f^+=\infty$ and $\int f^-=\infty$ – Lukaszmat Dec 31 '17 at 16:04
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    @Lukaszmat okay if that is the case, then with the function $g(x) $ we surely have $\int g^+, d\mu=\infty $ and $\int g^-, d\mu=\infty $. One can easily change the proofs given in the link I provided so that you get that the positive and the negative part both diverge to infinity. – Shashi Dec 31 '17 at 16:07
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    @Lukaszmat We use MCT to link the Lebesgue integral with the Riemann integral, just use $g_n(x)=\mathbf{1}{[0,n]}g^+(x)$ and get: \begin{align} \int{[0,\infty)} g^+(x),d\mu(x)=\int^\infty_0 g^+(x),dx \end{align} Moreover: \begin{align} \int^\infty_0 g^+(x),dx&=\sum_{k=0}^\infty \int^{(2k+1)\pi}{2k\pi} \frac{\sin(x)}{x},dx\ &\geq \sum{k=0}^\infty\frac{1}{(2k+1)\pi}\int^{(2k+1)\pi}{2k\pi} \sin(x),dx \ &= \sum{k=0}^\infty\frac{2}{(2k+1)\pi}\ &=\infty \end{align} The same can be done for $g^-$. So.... Conclude! – Shashi Dec 31 '17 at 16:45
  • Ok, now I understand :) – Lukaszmat Dec 31 '17 at 18:24