Question on Outer Expectation unequal to Expectation of minimal measurable majorant

Question

This is a problem given by Aad W. van der Vaart and Jon A. Wellner in their "weak convergence and stochastic process". It aims at constructing an example of "$E^{\ast}T$ is not always $ET^{\ast}$".

Let $(\Omega,\mathcal{A},P)$ be $\mathbb{R}$ with the $\sigma$-field generated by $[0,\infty)$ and the Borel sets in $(-\infty,0)$ and $P$ equal to the Cauchy measure. The map $T:\Omega\rightarrow\mathbb{R}$ defined by $T(\omega)=\omega$ has $E^{\ast}T=\infty$, but $ET^{\ast}$ does not exist.

Can someone help me to see why the statement is true?

For the definition of "Outer Expectation":

$E^{\ast}T=\inf\{EU:U\geq T,U:\Omega \rightarrow\bar{\mathbb{R}}\; \text{measurable and} \;EU \; \text{exists} \}$.

For the definition of "Minimal Measurable majorant":

For any map $T:\Omega\rightarrow\bar{\mathbb{R}}$, there exists a measurable function $T^{\ast}:\Omega \rightarrow \bar{\mathbb{R}}$ with (i) $T^{\ast}\geq T$; (ii) $T^{\ast}\leq U$ a.s., for every measurable $U:\Omega \rightarrow \bar{\mathbb{R}}$ with $U\geq T$ a.s.

Answer 1

(Standard) Cauchy distribution: (definition) Probability measure on $(\mathbb R,\mathcal B(\mathbb R))$ with density $f$ with respect to the Lebesgue measure, where $f(x)=\frac1{\pi(1+x^2)}$ for every $x$ in $\mathbb R$.

Answer 2

$T^*$ is just $T$, since it's already measurable, and $ET$ famously does not exist, so $ET^*$ does not exist.

But $E^*T$ has to exist: it's an infimum over a set of expectations that exist. So that at least shows $ET^*\neq E^*T$.

Now consider a majorant $U$ of $T$ and think about the expectation of the positive and negative part of $U$. The expectation of the positive part is infinite, so if $EU$ exists the expectation of the negative part must be finite, so $EU=\infty$, and since $U$ was arbitrary, $E^*T=\infty$