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Why $M^{2}_{[0,T]} \subset P^{2}_{[0,T]}$, where:

$M^{2}_{[0,T]} = \{f: \Omega \times [0, T] \to \mathbb{R}^{d \times m}: f - product \ measurable, (\mathcal{F_{t}})\ adapted, \mathbb{E} \int_{0}^{T} ||f(s)||^{2}ds< \infty\}$

$P^{2}_{[0,T]} = \{f: \Omega \times [0, T] \to \mathbb{R}^{d \times m}: f - product \ measurable, (\mathcal{F_{t}})\ adapted, \mathbb{P}( \int_{0}^{T} ||f(s)||^{2}ds< \infty ) =1\} ? $

I can't see why probability one does not result in finite expected value.

Thank you!

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This follows from the fact that if a random variable $Y$ is square integrable, then $P(Y^2<\infty)=1$ (essentially because $nP(Y^2>n)\leqslant \mathbb E\left[Y^2\mathbf 1\{Y^2>n\}\right]\leqslant \mathbb E\left[Y^2\right]$.

If $X$ is a random variable such that $\mathbb P(X^2<\infty)=1$ but $\mathbb E\left[X^2\right]$ is infinite, let $f\colon (\omega,t)\mapsto X(\omega)$. Then $f$ belongs to $P^2_{[0,T]}$ but not to $M^2_{[0,T]}$.

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  • $\begingroup$ But why there is no inclusion to left? What example of process is in $P^{2}$ but not in $M^{2}$? $\endgroup$ – tangerine Mar 26 '20 at 9:35
  • $\begingroup$ @tangerine I have added an example. $\endgroup$ – Davide Giraudo Mar 26 '20 at 10:28
  • $\begingroup$ What I missed earlier is the fact, that random variable cannot take infinity as a value. This post was also helpful: math.stackexchange.com/questions/239288/… $\endgroup$ – tangerine Mar 26 '20 at 12:32

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