Why $\int _0 ^t \phi_s ^2 ds < \infty \ \mathbb P \text{-a.e.}$ do not implies $\mathbb E [\int _0 ^t \phi_s ^2 ds] < \infty$?

Question

Why $\phi =(\phi_t)_{t \in [0,T]}$ is a progressive mesurable stochastic process do not implies $\mathbb E [\int _0 ^t \phi_s ^2 ds] < \infty$?

I know that if $X$ is a positive random variable such that $\mathbb E [X]< \infty$ then $X < \infty \ \mathbb P \text{-a.e.}$. However I don't remember how to prove it and neither why it's not equivalent (in other words, why does not the implication work in the inverse sense).

Could someone help me with this little doubt please ?

Answer 1

The converse is not true. The Cauchy random variable is finite a.s. but it does not have defined its mean. Same with the Pareto distribution for $\alpha \leq 1$ where the mean is infinite.

The property you mention comes from measure theory. If a function $f > 0$ is not finite a.e. then the integral of $f$ is already infinite over $\{f=\infty\}$. For general $f$ if also you have a non negligible set where is $-\infty$ then the integral is not defined.