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I've a stupid doubt in the construction of stochastic integral of real scalar valued maps. Many times I've seen in books after the stochastic integral is defined in [$0,T$] for the integrand in $L^2$ where $E(\int_{0}^{T}f(s)^2ds)<\infty$ we again use localization technique to extend the stochastic integral to more general processes $P(\int_{0}^{T}f(s)^2ds<\infty)=1 $. My questions are

  1. For a finite measure space if a random variable is almost everywhere finite then does not it mean that it expectation is finite? That too when I work in [$0,T$] not in [$0,\infty$]

  2. Can someone give me a counter example stating that a function belonging to the more general localization-technique defined space and not into the $L^2$ space.

I'm omitting the details as these are very standard and can be found in text books.

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up vote 5 down vote accepted

No, almost surely finite does not imply integrable. Consider a random variable $X$ exponential with parameter $1$ and $Y=\mathrm e^X$. Then $X$ is almost surely finite hence $Y$ is almost surely finite but, for every $y\geqslant1$, $\mathrm P(Y\geqslant y)=\mathrm P(X\geqslant\log y)=\mathrm e^{-\log y}=1/y$, hence $Y$ is not integrable.

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The chain is bounded P-a.s. -> integrable -> finite expectation -> finite P-a.s. right? – BCLC Feb 23 '15 at 6:37

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