Let $(\Omega,\mathcal A,\mathbb F,P)$ be a filtered complete probability space (i.e. $(\Omega,\mathcal A,P)$ is a complete probability space and $\Bbb F=(\mathcal F_t)_{\ge0}$ is a complete filtration wrt the given space).
Let's consider a process $X:\Omega\times[0,T]\to\Bbb R$ a.s. continous (i.e. $t\mapsto X_t(\omega)$ continous for almost all $\omega\in\Omega$) and $\Bbb F$-adapted. Thus $X\in M_{\bf{loc}}^2[0,T]$, thus $\int_0^TX_s^2\,ds<+\infty$ a.s..
I have to prove that for fixed $t_0\in[0,T]$, we have $\Bbb E[X_{t_0}^2]<+\infty$.
I tried as follows.
If $\Bbb E[X_{t_0}^2]=+\infty$, then for every $M>0$, there exist $A_M\in\mathcal F_{t_0}$ with $P(A_M)>0$, such that $X_{t_0}^2(\omega)>M$ for every $\omega\in A_M$.
Now by continuity, for fixed $\omega\in A_M$, there exists $r_{\omega}>0$ such that $$ X_{t}^2(\omega)>M\;\;\;\forall t\in]t_0-r_{\omega},t_0+r_{\omega}[\;\;. $$ Possibly shrinking $A_M$ (but always $P(A_M)>0$), we suppose $r:=\inf_{\omega\in A_M}r_{\omega}>0$, from which $$ X_t^2(\omega)>M\;\;\;\;\;\;\forall \omega\in A_M, t\in]t_0-r,t_0+r[ $$ thus $$ \int_0^TX_s^2\,ds\ge\int_{t_0-r}^{t_0+r}X_s^2\,ds\ge2rM\;\;. $$ The idea, is to deduce from this that $\int_0^TX_s^2\,ds$ cannot be finite and so reaching a contradiction.
The problem is that $r$ clearly depends on $M$, thus $rM$ could not be strictly increasing in $M$. I'm stucked. Can someone help me please? Thanks.
Without additional assumptions, the claim does, in general, not hold true.
Counterexample: Consider $\Omega = (0,1)$ endowed with Lebesgue measure $\lambda$ restricted to $(0,1)$. Define a process $(X_t)_{t \geq 0}$ by $$X_t(\omega) := \frac{1}{\omega}, \qquad \omega \in \Omega = (0,1),t \geq 0.$$ Since $t \mapsto X_t(\omega)$ is constant, it is obvious that $(X_t)_{t \geq 0}$ has continuous sample paths. However, $$\mathbb{E}(X_t^2) = \int_{(0,1)} \frac{1}{\omega^2} \, \lambda(d\omega) = \infty$$ for all $t \geq 0$.