Takesaki lemma: existence Gelfand-Pettis integral

Question

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 $\mathbb{R}$.

Thanks in advance for your help!

Answer 1

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.