# Some proof about Lebesgue integral of relaxing the restriction

There is some theorem stated on my textbook as following:

If either $\int_E f$ or $\int_{-\infty}^{+\infty}\alpha d\omega(\alpha)$ exists and is finite, then the other exists and is finite and $\int_E f=-\int_{-\infty}^{+\infty}\alpha d\omega(\alpha)$

,where $\int_E f$ is Lebesgue integral on measurable E and $\int_{-\infty}^{+\infty}\alpha d\omega(\alpha)$ is improper Riemann-Stieltjes and $\omega(\alpha)=\mu\bigg(\{x\in E :f(x)>\alpha\}\bigg),$ for each $\alpha\in \mathbb{R}$

I just wondering what happened if we relax the restriction of finiteness of integrals.So I modify the theorem as following :

If either $\int_E f$ or $\int_{-\infty}^{+\infty}\alpha d\omega(\alpha)$ exists , then the other exists and $\int_E f=\int_{-\infty}^{+\infty}\alpha d\omega(\alpha)$

Existence means that integrals need not be finite.

Is it true? If yes,can give some direction or hint to follow ?If no, is there any counterexample for it ?Thanks for considering my quest.

If we deal with the function $f(x)=1/x^{2}$ on $[0,1]$, then the Stieltjes integral does not exist if we interpret it in the regular sense:
For $0<a<M<\infty$, we have in some sense of integration by parts that \begin{align*} \int_{a}^{M}\alpha d\omega(\alpha)=M\omega(M)-a\alpha(a)-\int_{a}^{M}\omega(\alpha)d\alpha, \end{align*} simple calculations show that $M\omega(M)=M^{1/2}$ and $a\alpha(a)=a^{1/2}$, let $a\rightarrow 0^{+}$, the right side is then $M\omega(M)-\displaystyle\int_{0}^{M}\omega(\alpha)d\alpha$, where the latter integral is due to Monotone Convergence Theorem. However, we know that $\displaystyle\int_{0}^{\infty}\omega(\alpha)d\alpha=\int_{0}^{1}\dfrac{1}{x^{2}}dx=\infty$, where $\lim_{M\rightarrow\infty}\displaystyle\int_{0}^{M}\omega(\alpha)d\alpha=\int_{0}^{\infty}\omega(\alpha)d\alpha$ by Monotone Convergence Theorem again. Meanwhile, $M\omega(M)=M^{1/2}\rightarrow\infty$ as $M\rightarrow\infty$, so we end up with $\infty-\infty$ by the integration by parts whenever $a\rightarrow 0^{+}$ and $M\rightarrow\infty$, so $\lim_{M\rightarrow\infty,~a\rightarrow 0^{+}}\displaystyle\int_{a}^{M}\alpha d\omega(\alpha)$ does not exist even in the extended real sense.