# Takesaki lemma: existence Gelfand-Pettis integral

Consider the following fragment from Takesaki's second volume of "Theory of operator algebras" (lemma 2.4, chapter VI "Left Hilbert algebras"). In another post, it was explained that we should interpret the integral $$\int_{\mathbb{R}} e^{-rt^2}x(t)dt$$ as a Gelfand-Pettis integral.

Why does this Gelfand-Pettis integral exist? Is there a general existence result that garantuees the existence of this integral as en element of the compact subset $$K$$?

The closed I could find was theorem 3.27 in Rudin's book "Functional analysis" but there Rudin works with probability measures (or more generally, finite measures) on compact spaces, whereas here we are dealing with Lebesgue measure on $$\mathbb{R}$$.

• You can find a complete description (by double-duality) of this integral in Bourbaki Integration Chapter III § 3.1. Mar 21 at 5:47
• @DuchampGérardH.E. I quickly skimmed this section in Bourbaki's book but did not see anything in this section outside the case that the integrand function is compactly supported. Can you be a little more specific which result in this section is relevant? Thanks! Mar 21 at 7:17
• Okay I will send you some more elaborated route ASAP (when back, I'm travelling). Mar 21 at 7:57
• @DuchampGérardH.E. Thanks! Safe travels. Mar 21 at 8:08
• Does [ this ](:en.wikipedia.org/wiki/Pettis_integral) help? Mar 21 at 17:10

You can reduce your integral to the case treated by Rudin. Let $$\beta\mathbb R$$ be the Stone-Cech compactification of $$\mathbb R$$. By the universal property, $$x$$ extends to a continuous map $$\tilde x\colon\beta\mathbb R\to K$$.
Define a probability measure on $$B(\beta\mathbb R)$$ by $$\mu(A)=\sqrt{\frac{r}{\pi}}\int_{A\cap \mathbb R}e^{-rt^2}\,dt$$. Note that since the embedding $$\mathbb R\to \beta\mathbb R$$ is a homeomorphism onto its image, the trace $$\sigma$$-algebra of $$B(\beta\mathbb R)$$ on $$\mathbb R$$ coincides with $$B(\mathbb R)$$.
By Rudin's Theorem 3.27, the Pettis integral $$\int_{\beta\mathbb R}\tilde x\,d\mu$$ exists and belongs to $$K$$. Moreover, approximation by step functions one can show that $$\int_{\beta \mathbb R}\omega(\tilde x)\,d\mu=\sqrt{\frac{r}{\pi}}\int_{\mathbb R}\omega(x(t))e^{-rt^2}\,dt$$ for $$\omega\in E^\ast$$. Thus $$\int_{\beta\mathbb R}\tilde x\,d\mu=\sqrt{\frac{r}{\pi}}\int_{\mathbb R}x(t)e^{-rt^2}\,dt$$ as Pettis integrals.