This question is from a step in a proof in the book "Stochastic Calculus Applied to Finance" by D. Lamberton and B. Lapeyre, page 41.
Given an $\mathcal{F}_{t}$-adapted, mean square integrable process $H_{t}$, I want to prove that $\int_{0}^{t} H^{2}_{u} \, du$ is $\mathcal{F}_{t}$-measurable. The argument in the book is the following: "this result is true if $H$ is a simple process and, by density, it is true if $H$ is mean-square integrable".
I am stuck with the "by density" part. Indeed, given that the simple processes are dense in the space of mean square integrable processes with the norm $\|X\| = E[\int_{0}^{t} X^{2}_{u} \, du]$, choose a sequence $H^{n}$ of processes converging to $H$ in this space. One would like to prove that $\int_{0}^{t} (H^{n})^{2}_{u} \, du$, or any subsequence, converges to $\int_{0}^{t} H^{2}_{u} \, du$.
I have only figured out that $\mathbb{E}\big[\int_{0}^{t} (H_{s}^{n})^{2} \,\mathrm{d}s \big] - \mathbb{E}\big[\int_{0}^{t} H_{s}^{2} \, \mathrm{d}s \big] \xrightarrow[n\to \infty]{} 0$, but I need a convergence in $L^1$ of $\int_{0}^{t} (H_{s}^{n})^{2} \, \mathrm{d}s$ to $\int_{0}^{t} H_{s}^{2} \, \mathrm{d}s$ to finish the proof.
Since
$$|x^2-y^2| = |x+y| \cdot |x-y| \leq (|x|+|y|) |x-y|$$
for any $x,y \in \mathbb{R}$, we have
$$\begin{align*} \mathbb{E} \left| \int_0^t (H_s^n)^2 \, ds - \int_0^t H_s^2 \, ds \right| &= \mathbb{E} \int_0^t |(H_s^n)^2-H_s^2| \, ds \\ &\leq \mathbb{E} \left( \int_0^t |H_s-H_s^n| (|H_s| + |H_s^n|) \, ds \right). \end{align*}$$
Applying the Cauchy Schwarz inequality, we get
$$\begin{align*} \mathbb{E} \left| \int_0^t (H_s^n)^2 \, ds - \int_0^t H_s^2 \, ds \right| &\leq \sqrt{\mathbb{E} \left( \int_0^t |H_s-H_s^n|^2 \, ds \right)} \sqrt{\mathbb{E} \left( \int_0^t (|H_s|+|H_s^n|)^2 \, ds \right)} \\ &\leq \sqrt{2} \sqrt{\mathbb{E} \left( \int_0^t |H_s-H_s^n|^2 \, ds \right)} \sqrt{\mathbb{E} \left( \int_0^t (|H_s|^2+|H_s^n|^2) \, ds \right)} \end{align*}$$
where we have used that $$(x+y)^2 \leq 2x^2+2y^2.$$
As $H^n \to H$ in $L^2(\mathbb{P} \otimes \lambda|_{[0,t]})$, the right-hand side converges to $0$ as $n \to \infty$.