Interchangability of expected value and integral only when $\mathbb E|X(t)| < \infty$ for a stochastic process $\{ X(t) \}$?

In my textbook of Stochastic Process (in Chinese), I found a theorem as follows:

Theorem: If a stochastic process $\{ X(t), t \ge 0 \}$ has $\mathbb E|X(t)| < \infty$ and $\mathrm{var} X(t) < \infty$, then$$\mathbb E \left[ \int_0^t X(s) \, ds \right] = \int_0^t \mathbb E [ X(s) ] \, ds, \\ \mathbb E\left[ \int_0^s \int_0^t X(v) X(u) \, du dv \right] = \int_0^s \int_0^t \mathbb E[X(v)X(u)] \, du dv$$

My textbook says this can be simply proved by applying Fubini's theorem.

But I am confused. To apply Fubini's theorem, we seem to need a condition like $\int_0^t \mathbb E |X(t)| < \infty$ as in this. Only $\mathbb E|X(t)| < \infty$ is really enough?

I doubt this, but I have no idea how to construct a counterexample. So I ask here, thanks for any help!