Given a regular (that is, right continuous with left limits) adapted stochastic process $(X_t)_{t\geq 0}$ and given that $\int _0 ^{T} \mathbf{E}\left[X_s^2\right] < +\infty$ how does one prove that $\int _0^T X_s^2 ds < +\infty$?
My attempt:
Using Fubini's theorem (whose hypothesis are indeed satisfied) we see that $\mathbf{E}\left[\int _0 ^{T} X_s^2\right]ds = \int _0 ^{T} \mathbf{E}\left[X_s^2\right]ds < +\infty$ , but how do I proceed further? Can I conclude that since the expectation is finite, the random variable will be finite?
This is really elementary and thus doesn't require Jensen's inequality.
Lemma. If $Y$ is a $\mathbf{P}$-a.s. non-negative random variable such that $\mathbf{E}\left[Y\right] < + \infty$, then $Y < +\infty$ $\mathbf{P}$-a.s.
Proof. Indeed, suppose that we don't have $Y < +\infty$ $\mathbf{P}$-a.s. This means that $\mathbf{P}\left[\{ Y < + \infty \}\right] < 1$ hence that $\mathbf{P}\left[\{ Y = + \infty \}\right] > 0$. Noting $A := \{ Y = + \infty \}$ we then see that $\mathbf{E}\left[Y\right] = \int_{\Omega} Y d\mathbf{P} \geq \int_A Y d\mathbf{P} = \int_{A} (+\infty) d\mathbf{P} = +\infty$ which contradicts the fact that $\mathbf{E}\left[Y\right] < + \infty$. (For the last equality remark that $\int_{A} (+\infty) d\mathbf{P} \geq \int_{A} n d\mathbf{P} = n \mathbf{P}\left[A\right]$ for all $n\in\mathbf{N}$ and use the fact that $\mathbf{P}\left[A\right] > 0$.)
Applying the lemma to $Y = \int_0^T X_s^2 ds$ allows to conclude.