Proof of Linearity, Mean Zero, and Isometry of the Elementary Stochastic Integral

Step 0 (Common representation). By the refinement argument recorded in Elementary Stochastic Integral of a Simple Adapted Process, we may represent $H$ and $G$ over one common partition $0=t_0<t_1<\dots<t_n=T$ with coefficients $\xi_i$ and $\zeta_i$: on a refinement interval $(u_j,u_{j+1}]\subseteq(t_i,t_{i+1}]$ of the original representation of $H$, the coefficient $\xi_i$ is $\mathcal{F}_{t_i}$-measurable and $\mathcal{F}_{t_i}\subseteq\mathcal{F}_{u_j}$ by the filtration property, so the refined data is again a representation in the sense of Simple Adapted Process, and the elementary integrals are unchanged. Write $\Delta_i=M_{t_{i+1}}-M_{t_i}$ and

using clause (iv) of Ito Integrator of Intensity Type.

Step 1 (Preliminary facts). Each $\Delta_i$ is square-integrable (difference of square-integrable random variables, by the closure properties of Square-Integrable Random Variables and the Mean-Square Inner Product). Its expectation vanishes: by the constant-expectation property of a square-integrable martingale, $\mathbb{E}[M_t]=\mathbb{E}[M_0]$ for all $t$, and $|\mathbb{E}[M_0]|=|\mathbb{E}[M_0-0]|\le\lVert M_0\rVert_2=0$ by the Cauchy-Schwarz inequality (with the constant $1$) and the null-equivalence clause of Square-Integrable Random Variables and the Mean-Square Inner Product, since $M_0=0$ almost surely by clause (ii) of Ito Integrator of Intensity Type. Hence $\mathbb{E}[\Delta_i]=0$.

Let $\eta$ be any $\mathcal{F}_{t_i}$-measurable square-integrable random variable. 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, $\eta$ and $\Delta_i$ are independent; moreover $\eta^{2}$ and $\Delta_i^{2}$ are independent, since $\sigma(\eta^{2})\subseteq\sigma(\eta)$ and $\sigma(\Delta_i^{2})\subseteq\sigma(\Delta_i)$ (composition with the Borel function $x\mapsto x^{2}$, as in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras) and sub-$\sigma$-algebras of independent $\sigma$-algebras are independent directly from the definition. Both $\eta^{2}$ and $\Delta_i^{2}$ have finite expectation, so Expectation of a Product of Independent Random Variables gives

In particular $\eta\Delta_i$ is square-integrable.

Step 2 (Linearity). With the common representation, $aH+bG$ has representation $\bigl((t_i),(a\xi_i+b\zeta_i)\bigr)$: the coefficients are $\mathcal{F}_{t_i}$-measurable (linear combinations of measurable functions are measurable, by the rational-decomposition argument recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process applied on the measurable space $(\Omega,\mathcal{F}_{t_i})$) and square-integrable by the closure properties of Square-Integrable Random Variables and the Mean-Square Inner Product. Hence $aH+bG$ is a simple adapted process, and

which is claim 1.

Step 3 (Square-integrability and mean zero). $\int_0^T H_t\,dM_t=\sum_i\xi_i\Delta_i$ is a finite sum of square-integrable random variables (Step 1 with $\eta=\xi_i$), hence square-integrable by Square-Integrable Random Variables and the Mean-Square Inner Product, and by linearity of the expectation (Linearity and Monotonicity of the Lebesgue Integral) together with (1), $\mathbb{E}[\sum_i\xi_i\Delta_i]=\sum_i\mathbb{E}[\xi_i\Delta_i]=0$. This is claim 2.

Step 4 (Vanishing cross terms). Fix $i<j$. The random variable $\xi_i\zeta_j\Delta_i$ is $\mathcal{F}_{t_j}$-measurable: $\xi_i$ and $\zeta_j$ are $\mathcal{F}_{t_j}$-measurable (filtration monotonicity), $\Delta_i=M_{t_{i+1}}-M_{t_i}$ is $\mathcal{F}_{t_j}$-measurable because $M$ is adapted and $t_{i+1}\le t_j$, and products and differences of measurable functions are measurable (arguments recorded in Square-Integrable Random Variables and the Mean-Square Inner Product and Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process). It is also integrable: $\xi_i\Delta_i$ and $\zeta_j$ are square-integrable, and the product of two square-integrable random variables is integrable by Square-Integrable Random Variables and the Mean-Square Inner Product. By clause (iii) of Ito Integrator of Intensity Type and Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, $\xi_i\zeta_j\Delta_i$ and $\Delta_j$ are independent, both with finite expectation, so Expectation of a Product of Independent Random Variables and Step 1 give

By symmetry the same holds for $j<i$ with the roles of the factors exchanged.

Step 5 (Diagonal terms and the isometry). For each $i$, the product $(\xi_i\Delta_i)(\zeta_i\Delta_i)=(\xi_i\zeta_i)\Delta_i^{2}$ is integrable (product of the square-integrable random variables $\xi_i\Delta_i$ and $\zeta_i\Delta_i$). The factors $\xi_i\zeta_i$ (integrable, $\mathcal{F}_{t_i}$-measurable) and $\Delta_i^{2}$ (integrable, with $\sigma(\Delta_i^{2})\subseteq\sigma(\Delta_i)$) are independent as in Step 1, so Expectation of a Product of Independent Random Variables gives

Expanding the product of the two finite sums and using linearity of expectation with (2) and (3),

On the other hand, $t\mapsto\mathbb{E}[H_tG_t]$ equals the constant $\mathbb{E}[\xi_i\zeta_i]$ on $(t_i,t_{i+1}]$, so its extension by $0$ is the finite sum $\sum_i\mathbb{E}[\xi_i\zeta_i]\mathbf{1}_{(t_i,t_{i+1}]}$, a measurable step function. Its product with $\rho$ is integrable, being dominated in absolute value by $C\,\mathbf{1}_{(0,T]}\rho$ with $C=\max_i|\mathbb{E}[\xi_i\zeta_i]|$ (finiteness of $\int\mathbf{1}_{(0,T]}\rho\,d\lambda$ is clause (iv) of Ito Integrator of Intensity Type), and by linearity of the Lebesgue integral (Linearity and Monotonicity of the Lebesgue Integral),

Combining the two displays proves the polarization identity, and $G=H$ gives the isometry. $\square$