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 &lt; ∞ for all t ≥ 0? The question is clearly trivial, when ( X t ) t ≥ 0 and f are continuous.

Question

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        &amp;lt;        ∞            for all   t  ≥  0? The question is clearly trivial, when   (      X    t        )          t      ≥      0       and   f are continuous.

Myla Pierce · Accepted Answer

Example:   E  =      R  ,       X    s    =  1      /    (  1  −  s  ) if   0  ≤  s  &amp;lt;  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 &quot;right continuous with left limits&quot;. Suppose that by &quot;locally bounded&quot; you mean that       |    f      |   is bounded on each metric ball       B    r    (  x  )  :=  {  y  ∈  E  :  d  (  x  ,  y  )  &amp;lt;  r  }. Fix   t  &amp;gt;  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.