 tr2os8x

2022-06-28

Let $E$ be a metric space, $\left({X}_{t}{\right)}_{t\ge 0}$ be an $E$-valued right-continuous process and $f:E\to \left[0,\mathrm{\infty }\right)$ be locally bounded and Borel measurable. Is this enough to ensure that
$\begin{array}{}\text{(1)}& {\int }_{0}^{t}f\left({X}_{s}\right)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}s<\mathrm{\infty }\end{array}$
for all $t\ge 0$? The question is clearly trivial, when $\left({X}_{t}{\right)}_{t\ge 0}$ and $f$ are continuous. Myla Pierce

Expert

Example: $E=\mathbb{R}$, ${X}_{s}=1/\left(1-s\right)$ if $0\le s<1$, and $=1$ if $s\ge 1$. This (non-random) path is right continuous, but with $f\left(x\right):=|x|$ (certainly locally bounded) the integral ${\int }_{0}^{2}f\left({X}_{s}\right)\phantom{\rule{thinmathspace}{0ex}}ds$ diverges.
Fix: Strengthen the hypothesis on $X$ to "right continuous with left limits". Suppose that by "locally bounded" you mean that $|f|$ is bounded on each metric ball ${B}_{r}\left(x\right):=\left\{y\in E:d\left(x,y\right). Fix $t>0$ and $\omega \in \mathrm{\Omega }$. The real-valued function $s↦d\left({X}_{0}\left(\omega \right),{X}_{s}\left(\omega \right)\right)$ is then right continuous with left limits for $s\in \left[0,t\right]$. As such, it is bounded. Therefore there is a constant $C=C\left(\omega ,t\right)$ such that $0\le f\left({X}_{s}\left(\omega \right)\right)\le C\left(t,\omega \right)$ for all $s\in \left[0,t\right]$. The integral ${\int }_{0}^{t}f\left({X}_{s}\left(\omega \right)\right)\phantom{\rule{thinmathspace}{0ex}}ds$ therefore converges.

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