Let E be a metric space, ( X t ) t ≥ 0 be an...

tr2os8x

tr2os8x

Answered

2022-06-28

Let E be a metric space, ( X t ) t 0 be an E-valued right-continuous process and f : E [ 0 , ) be locally bounded and Borel measurable. Is this enough to ensure that
(1) 0 t f ( X s ) d s <
for all t 0? The question is clearly trivial, when ( X t ) t 0 and f are continuous.

Answer & Explanation

Myla Pierce

Myla Pierce

Expert

2022-06-29Added 20 answers

Example: E = R , X s = 1 / ( 1 s ) if 0 s < 1, and = 1 if s 1. This (non-random) path is right continuous, but with f ( x ) := | x | (certainly locally bounded) the integral 0 2 f ( X s ) d s 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 ( x ) := { y E : d ( x , y ) < r }. Fix t > 0 and ω Ω. The real-valued function s d ( X 0 ( ω ) , X s ( ω ) ) is then right continuous with left limits for s [ 0 , t ]. As such, it is bounded. Therefore there is a constant C = C ( ω , t ) such that 0 f ( X s ( ω ) ) C ( t , ω ) for all s [ 0 , t ]. The integral 0 t f ( X s ( ω ) ) d s therefore converges.

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