# 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 ... [closed]

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?

• Is it a Riemann integral of a Lebesgue one?
– user515010
Commented Dec 31, 2017 at 14:26
• Is is a Lebesgue. Commented Dec 31, 2017 at 14:30

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.
• Yes, but maybe this integral exists and is equal $\infty$ or $-\infty$ Commented Dec 31, 2017 at 15:48
• 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. Commented Dec 31, 2017 at 15:56
• For me does not exist means that $\int f^+=\infty$ and $\int f^-=\infty$ Commented Dec 31, 2017 at 16:04
• @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. Commented Dec 31, 2017 at 16:07
• @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! Commented Dec 31, 2017 at 16:45