Almost-sure finiteness does not imply integrability

By One-dimensional Brownian motion hits every point almost surely, ${\tau }_a$ is a measurable $[0,\mathrm{\infty }]$-valued random time and the measurable event $N:=\{{\tau }_a=+\mathrm{\infty }\}$ has probability zero.

Define $T_a(\omega )={\tau }_a(\omega )$ for $\omega \mathrm{\notin }N$ and $T_a(\omega )=0$ for $\omega \in N$. Then $T_a$ takes values in $[0,\mathrm{\infty })\subset \mathbb{R}$ on every outcome. It is measurable: for $c<0$ the set $\{T_a\le c\}$ is empty, while for $c\ge 0$ it is $N\cup \{{\tau }_a\le c\}$. Thus $T_a$ is a real random variable in the sense of Random elements and real random variables, and in particular a real-valued random time.

The variables $T_a$ and ${\tau }_a$ differ only on the null event $N$. Consequently, for every Borel set $C\subseteq [0,\mathrm{\infty })$, the events $\{T_a\in C\}$ and $\{{\tau }_a\in C\}$ have symmetric difference contained in $N$, so they have the same probability. Hence $T_a$ has the density $a(2\pi t^3{)}^{-1\mathrm{/}2}{e}^{-a^2\mathrm{/}(2t)}$ from A Brownian hitting time has infinite mean, and the same supplier's calculation gives $E[T_a]=+\mathrm{\infty }$.

Therefore $T_a$ is finite on every outcome, hence finite almost surely, but is not integrable. This real-valued witness refutes the stated implication. The null-set modification is explicit, and AC is used only through the Brownian construction and the two cited hitting-time suppliers.