$\def\R{\mathbb{R}}\def\ndot{{\,\cdot\,}}\def\d{\mathrm{d}}\def\Ω{{\mit Ω}}\def\peq{\mathrel{\phantom{=}}{}}$Lema: Supongamos que $(U, \mathscr{F}, μ)$ e $(V, \mathscr{G}, ν)$ son medibles espacios. Si $f: U × V → \R$ es $\mathscr{F} × \mathscr{G}$medible función tal que$$
\int_U |f(x, y)| μ(\d x) < +∞
$$
para casi todas las $y \in V$, luegode$$
\left( \int_V \left( \int_U f(x, y) µ(\d x) \right)^2 ν(\d y) \right)^{\tfrac{1}{2}} \leqslant \int_U \left( \int_V (f(x, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} µ(\d x).
$$
Prueba: Por la de Cauchy-Schwarz desigualdad,\begin{align*}
\text{RHS}^2 &= \left( \int_U \left( \int_V (f(x, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x) \right)^2\\
&= \left( \int_U \left( \int_V (f(x_1, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x_1) \right) \left( \int_U \left( \int_V (f(x_2, y))^2 ν(\d y) \right)^{\smash{\tfrac{1}{2}}} μ(\d x_2) \right)\\
&= \iint_{U^2} \left( \int_V (f(x_1, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} \left( \int_V (f(x_2, y))^2 ν(\d y) \right)^{\tfrac{1}{2}} μ(\d x_1) μ(\d x_2)\\
&\geqslant \iint_{U^2} \left( \int_V f(x_1, y) f(x_2, y) ν(\d y) \right) μ(\d x_1) μ(\d x_2)\\
&= \int_V \left( \iint_{U^2} f(x_1, y) f(x_2, y) μ(\d x_1) μ(\d x_2) \right) ν(\d y)\\
&= \int_V \left( \int_U f(x_1, y) μ(\d x_1) \right) \left( \int_U f(x_2, y) μ(\d x_2) \right) ν(\d y)\\
&= \int_V \left( \int_U f(x, y) μ(\d x) \right)^2 ν(\d y) = \text{LHS}^2.
\end{align*}
Ahora volviendo a la pregunta y supongo que $X: [0, T] × \Ω → \R$ es un medibles proceso tal que$$
\int_0^T E(|X_t|) \,\d t < +∞. \etiqueta{$*$}
$$
Desde$$
\int_\Ω \int_0^T |X_t(ω)| \,\d t \d ω = \int_0^T \int_\Ω |X_t(ω)| \,\d t \d ω = \int_0^T E(|X_t|) \,\d t < +∞,
$$
a continuación, $\displaystyle \int_0^T |X_t| \,\d t < +∞$ casi seguramente, $\displaystyle E\left( \int_0^T X_t \,\d t \right) = \int_0^T E(X_t) \,\d t$ existe, y por lo tanto $E(X_t)$ existe para casi todos los $t \in [0, T]$. Para tal $t$, definir $m(t) = E(X_t)$, $Y_t = X_t - m(t)$, a continuación, $D(X_t) = E(Y_t^2)$ e$$
E\left( \int_0^T Y_t \,\d t \right) = E\left( \int_0^T X_t \,\d t \right) - \int_0^T m(t) \,\d t = 0,
$$
lo que implica$$
D\left( \int_0^T X_t \,\d t \right) = D\left( \int_0^T Y_t \,\d t + \int_0^T m(t) \,\d t \right) = D\left( \int_0^T Y_t \,\d t \right) = E\left( \left( \int_0^T Y_t \,\d t \right)^2 \right).
$$
Por el lema,$$
\left( D\left( \int_0^T X_t \,\d t \right) \right)^{\tfrac{1}{2}} = \left( E\left( \left( \int_0^T Y_t \,\d t \right)^2 \right) \right)^{\tfrac{1}{2}} \leqslant \int_0^T \sqrt{\smash[b]{E(Y_t^2)}} \,\d t = \int_0^T \sqrt{\smash[b]{D(X_t)}} \,\d t.
$$
