Proof of The Elementary Stochastic Integral Process is a Square-Integrable Martingale

Fix a representation $\bigl((t_i)_{i=0}^{n},(\xi_i)_{i=0}^{n-1}\bigr)$ of $H$ as in Simple Adapted Process.

Step 1 (Restrictions are simple adapted). Fix $t\in(0,T]$ and let $i^{*}$ be the largest index with $t_{i^{*}}<t$. Then $0=t_0<t_1<\dots<t_{i^{*}}<t$ is a partition of $[0,t]$, and on $(t_i,\min(t_{i+1},t)]$ the restriction of $H$ takes the value $\xi_i$; so $\bigl((t_0,\dots,t_{i^{*}},t),(\xi_0,\dots,\xi_{i^{*}})\bigr)$ is a representation of $(H_u)_{u\in(0,t]}$ on $(0,t]$, which is therefore a simple adapted process, and by Elementary Stochastic Integral of a Simple Adapted Process

Step 2 (Adaptedness and square-integrability). In (1), each $\xi_i$ is $\mathcal{F}_{t_i}$-measurable with $t_i<t$, and each value of $M$ is taken at a time $\le t$, so it is $\mathcal{F}_t$-measurable because $M$ is adapted and the filtration is increasing. Sums, differences, and products of measurable functions are measurable (the arguments recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process and Square-Integrable Random Variables and the Mean-Square Inner Product, applied on $(\Omega,\mathcal{F}_t)$), so $I_t$ is $\mathcal{F}_t$-measurable; and $I_t$ is square-integrable by claim 2 of Linearity, Mean Zero, and Isometry of the Elementary Stochastic Integral applied on $(0,t]$. Also $I_0=0$ is measurable for every $\sigma$-algebra and square-integrable. Consequently $\tilde I_s=I_{\min(s,T)}$ is $\mathcal{F}_{\min(s,T)}$-measurable, hence $\mathcal{F}_s$-measurable, and square-integrable, for every $s\ge0$: the process $(\tilde I_s)_{s\ge0}$ is adapted with square-integrable values.

Step 3 (Averaged martingale identity on $[0,T]$). Let $0\le s\le t\le T$ and $A\in\mathcal{F}_s$; we may assume $s<t$. Refining the representation of $H$ by inserting the points $s$ and $t$ (as in Elementary Stochastic Integral of a Simple Adapted Process, which leaves all elementary integrals unchanged), we may assume $s$ and $t$ are partition points, say $s=t_{a}$ and $t=t_{b}$ with $a\le b$ (with the convention $s=t_0=0$ when $s=0$). Then by (1),

Fix $i\in\{a,\dots,b-1\}$. The random variable $\xi_i\mathbf{1}_{A}$ is $\mathcal{F}_{t_i}$-measurable ($A\in\mathcal{F}_s\subseteq\mathcal{F}_{t_i}$, and the product of measurable functions is measurable) and square-integrable ($|\xi_i\mathbf{1}_A|\le|\xi_i|$ pointwise, with monotonicity from Linearity and Monotonicity of the Lebesgue Integral). By clause (iii) of Ito Integrator of Intensity Type and the final paragraph of Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, $\xi_i\mathbf{1}_A$ and $\Delta_i$ are independent; both are integrable (square-integrable random variables are integrable by Square-Integrable Random Variables and the Mean-Square Inner Product), so Expectation of a Product of Independent Random Variables gives

where $\mathbb{E}[\Delta_i]=0$ as in Step 1 of the proof of Linearity, Mean Zero, and Isometry of the Elementary Stochastic Integral (constant expectation of the square-integrable martingale $M$ together with $M_0=0$ almost surely). Summing over $i$ and using linearity of expectation (Linearity and Monotonicity of the Lebesgue Integral),

Step 4 (The stopped extension is a martingale). Let $0\le s\le t$ be arbitrary and $A\in\mathcal{F}_s$. If $s\ge T$, then $\tilde I_t-\tilde I_s=I_T-I_T=0$ and the averaged identity is trivial. If $s<T$, then $\tilde I_t-\tilde I_s=I_{\min(t,T)}-I_s$ with $s\le\min(t,T)\le T$, and (2) applies. Hence

which together with Step 2 is exactly the averaged form of the martingale property in Square-Integrable Martingale, Submartingale, and Supermartingale; by the equivalence recorded there, $(\tilde I_s)_{s\ge0}$ is a square-integrable martingale. $\square$