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!
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 $M^2_{[0,T]}$.