Continuous Semimartingales and the Monotone-Class Argument
The construction of a stochastic integral repeatedly uses two reductions. An identity is first checked for elementary objects and then extended to a measurable class; an estimate is first proved under boundedness and then extended by stopping. This post gives the details of both reductions, including the construction of quadratic variation and the passage from square-integrable martingales to continuous semimartingales.
Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.
To prove an identity for all measurable sets or functions, it often suffices to prove it on a generating class and check closure under suitable limits. The monotone-class argument makes this reduction precise. Let \(E\) be an arbitrary set, and let \(\mathcal P(E)\) denote the set of all subsets of \(E\). If \(C \subset \mathcal P(E)\), then \(\sigma(C)\) means the smallest \(\sigma\)-field on \(E\) containing \(C\) (which is equivalently understood as the intersection of all \(\sigma\)-fields containing \(C\)).
We say \(\mathcal M \subset \mathcal P(E)\) is a monotone class if
\(E \in \mathcal M\);
if \(A,B \in \mathcal M\) such that \(A \subset B\), then \(B \setminus A \in \mathcal M\); and
if \(\{A_n\}_{n\geq 0}\) is an increasing sequence of sets such that \(A_n \in \mathcal M\) for all \(n\), then \(\bigcup_{n \geq 0} A_n \in \mathcal M\).
At first glance, it is clear that a \(\sigma\)-field is a monotone class. We can also define the monotone class generated by \(C\) (which is an arbitrary collection of subsets of \(E\)), denoted \(\mathcal M(C) \subset \mathcal P(E)\), as
Theorem 1 (Monotone Class Theorem). If \(C \subset \mathcal P(E)\) is stable under finite intersections, then \(\mathcal M(C) = \sigma(C)\).
Proof. First note that any \(\sigma\)-field is a monotone class, so we have \(\mathcal M(C) \subset \sigma(C)\). For the reverse inclusion, the task is to prove closure under finite intersections; the defining closure properties then supply complements and countable unions. Note that this can be shown if we establish \(\mathcal M(C)\) is a \(\sigma\)-field. However, a monotone class is a \(\sigma\)-field if and only if it is stable under finite intersections: if we can show this, then, taking the complement, the collection of sets must then be stable under finite unions. Then, by definition, we can pass any finite union to the limit, and get that the complement of these sets is also stable under countable unions, which would then verify the proposition.
Write \(\mathcal D=\mathcal M(C)\). First fix \(A\in C\) and define \(\mathcal D_A=\{B\in\mathcal D:A\cap B\in\mathcal D\}\). This is a monotone class in the convention above (also called a Dynkin system): it contains \(E\), is closed under nested differences, and is closed under increasing unions. For example, if \(B\subset B'\), then \(A\cap(B'\setminus B)=(A\cap B')\setminus(A\cap B)\). Since \(C\) is intersection-stable, \(C\subset\mathcal D_A\). Minimality gives \(\mathcal D_A=\mathcal D\). Thus intersections of one member of \(C\) with one member of \(\mathcal D\) lie in \(\mathcal D\).
Now fix \(A\in\mathcal D\). The same closure argument applies to \(\mathcal D_A\), and the first stage, with the factors interchanged, gives \(C\subset\mathcal D_A\). Hence \(\mathcal D_A=\mathcal D\) again. This proves intersection stability for two arbitrary members of \(\mathcal D\), so \(\mathcal D\) is a sigma-field and equals \(\sigma(C)\). ◻
1. Continuous Semimartingales
Theorem 2. Let \(M\) be a continuous local martingale with \(M_0=0\). There is a unique continuous adapted increasing process \(A\), with \(A_0=0\), such that \(M^2-A\) is a continuous local martingale. We write \(A=\langle M\rangle\). For every deterministic sequence of partitions \(\pi_n\) of \([0,T]\) whose mesh tends to zero,
For a nonzero integrable initial value, define \(\langle M\rangle=\langle M-M_0\rangle\). The corresponding centered-square characterization always holds. The uncentered assertion \(M^2-\langle M\rangle\) is a local martingale under our integrable-initial-value convention if \(M_0\in L^2\).
We give a construction from martingale increments, before using stochastic integration. Suppose first \(M_0=0\) and \(|M_t|\leq D\). For a finite partition \(\pi\) of \([0,T]\), set
Each summand is a martingale. After time \(u\), its coefficient is measurable; before time \(u\), condition first at \(u\), where the expected future increment is zero. Thus \(X^\pi\) is a bounded continuous martingale.
Lemma 3. As the meshes of \(\pi\) and \(\sigma\) tend to zero, \(\mathbb{E}|X_T^\pi-X_T^\sigma|^2\to0\).
Take the common refinement \(\rho\). The difference is a martingale transform \(\sum_jD_j\Delta M_j\), where \(D_j\) is the difference of the two relevant left-endpoint values and is measurable at the left endpoint of the refined interval. Orthogonality of martingale increments gives
The first factor tends to zero by path continuity and bounded convergence. To bound the second uniformly, expand \(Q_\rho^2\). The fourth-power terms are at most \(4D^2\mathbb{E} Q_\rho\). Conditional orthogonality gives \(\mathbb{E}[\sum_{k>j}(\Delta M_k)^2\mid\mathcal F_{t_j}]=\mathbb{E}[(M_T-M_{t_j})^2\mid\mathcal F_{t_j}]\leq4D^2\). Hence
This proves the lemma for arbitrary partitions, without a nesting assumption.
Doob’s inequality now makes \(X^{\pi_n}\) Cauchy in the norm \(\bigl(\mathbb{E}\sup_{t\leq T}|X_t|^2\bigr)^{1/2}\). A subsequence with summable expected uniform differences converges uniformly almost surely; its limit \(Y\) is continuous, adapted, and a martingale, and the full sequence converges in that norm. Set \(A_t=M_t^2-2Y_t\). At partition points, the telescoping identity is
The right side is increasing in \(i\). Approximate any two fixed times by partition points, use uniform convergence and continuity, and then intersect over rational times. This proves that \(A\) is continuous and increasing, with \(A_0=0\). Moreover, \(M^2-A=2Y\) is a martingale. At \(T\), the same identity gives convergence of square sums in \(L^2\).
Uniqueness uses the elementary fact that a continuous local martingale of finite variation, starting at zero, vanishes. Stop such a process \(N\) when either its absolute value or its total variation reaches an integer, and at an integer time. For the stopped bounded martingale and a partition of \([0,T]\), orthogonality gives
Continuity and the stopped bounds justify dominated convergence. Removing the stops proves \(N=0\). The difference of two candidate compensators has these properties, proving uniqueness. It also makes constructions on different finite horizons consistent.
For a general zero-start continuous local martingale, use \(\tau_n=\inf\{t\geq0:|M_t|\geq n\}\wedge n\). The bounded stopped martingales have brackets that agree on overlaps by uniqueness. They therefore define a continuous increasing bracket for \(M\). On \(\{\tau_n>T\}\), square sums and brackets agree with those of the stopped process; these events have probability tending to one. This gives the asserted convergence in probability. For nonzero initial values, center first. If \(M_0\in L^2\), then \(M_0^2+2M_0(M-M_0)\) is a local martingale, by Example 7 below, which justifies the uncentered formula.
Lemma 4. Let \(M=\{M_t\}_{t\geq0}\) be a continuous local martingale and let \(T\) be a stopping time. Then, for every \(t\), \(\langle M^T, M^T \rangle_t = \langle M, M \rangle_{t \wedge T}\) almost surely.
Proof. By the centered-square characterization, \((M-M_0)^2-\langle M\rangle\) is a continuous local martingale. Stopping at \(T\) gives \((M_{t\wedge T}-M_0)^2-\langle M\rangle_{t\wedge T}\). The stopped bracket is continuous, adapted, increasing, and zero initially. Uniqueness in Theorem 2, applied to \(M^T-M_0\), therefore identifies it with \(\langle M^T\rangle\). This argument requires no second moment for \(M_0\). ◻
Lemma 5. Let \(M\) be a continuous local martingale such that \(M_0 = 0\). Then, \(\langle M, M \rangle = 0\) if and only if \(M \equiv 0\).
Proof. Suppose \(\langle M,M \rangle =0\). Then, \(M_t^2\) is a nonnegative continuous local martingale, hence it is a supermartingale, so by definition \(\mathbb{E}[M_t^2] \leq \mathbb{E}[M_0^2] = 0\). Therefore, \(M \equiv 0\). The converse is obviously true. ◻
Two continuous local martingales \(M\) and \(N\) are said to be orthogonal if \(\langle M,N\rangle=0\). By polarization of the centered-square characterization, this is equivalent to \((M-M_0)(N-N_0)\) being a continuous local martingale: the difference between this product and its covariation is a local martingale, and a continuous local martingale of finite variation starting at zero vanishes. If \(M_0N_0\in L^1\), the equivalent condition can also be written with the uncentered product \(MN\), because the two cross terms \(M_0(N-N_0)\) and \(N_0(M-M_0)\) are local martingales by Example 7. The initial-product condition is necessary for that formulation. In particular, two independent Brownian motions starting at zero are orthogonal. The Kunita–Watanabe inequality is the corresponding Cauchy–Schwarz bound for integrals against covariation.
Theorem 6 (Kunita-Watanabe; Le Gall (2016, Proposition 4.18)). Let \(M\) and \(N\) be two continuous local martingales, and \(H\) and \(K\) be two measurable processes. Then,
almost surely.
To verify the inequality, define \(C=\langle M,N\rangle=(\langle M+N\rangle-\langle M-N\rangle)/4\). The increment-square limits show that \(d\langle aM+bN\rangle=a^2d\langle M\rangle+2ab\,dC+b^2d\langle N\rangle\) is a positive measure for every rational \(a,b\), outside one null set. With respect to \(\mu=d\langle M\rangle+d\langle N\rangle\), the density matrix \(\begin{pmatrix}m&c\\c&n\end{pmatrix}\) is therefore positive semidefinite almost everywhere, so \(|c|\leq\sqrt{mn}\). Applying Cauchy–Schwarz to \(|H|\sqrt m\) and \(|K|\sqrt n\) proves the stated inequality, first on finite intervals and then by monotone convergence.
Example 7. If \(Z\) is a finite \(\mathcal F_0\)-measurable random variable and \(M\) is a continuous local martingale with \(M_0=0\), then \(ZM\) is a continuous local martingale. Choose a reducing sequence \(\sigma_n\) for \(M\). Set \(\tau_n=\sigma_n\) on \(\{|Z|\leq n\}\) and \(\tau_n=0\) otherwise. These are increasing stopping times tending to infinity. The process
is a uniformly integrable martingale, because its multiplier is bounded and \(\mathcal F_0\)-measurable. This proves the claim. If \(M_0\ne0\), apply it to \(M-M_0\) and additionally require \(ZM_0\in L^1\) in order to include the constant initial process. One cannot discard this integrability condition by saying that centering is without loss.
Example 8. A useful tail argument bounds moments of quadratic variation. Let \(A\) be continuous, adapted and increasing, with \(A_0=0\), \(A_\infty<\infty\) almost surely. Suppose \(Z\geq0\) is integrable and, for every stopping time \(T\),
where \(A_T=A_\infty\) when \(T=\infty\). At \(T_\lambda=\inf\{t:A_t>\lambda\}\), continuity gives \(A_{T_\lambda}=\lambda\) on \(\{A_\infty>\lambda\}\). Therefore
For an increasing \(C^1\) function \(f\geq0\) with \(f(0)=0\), put \(F(x)=\int_0^xf(y)\,dy\). The nonnegative identity \(F(x)=\int_0^\infty(x-\lambda)_+f'(\lambda)\,d\lambda\) and Tonelli’s theorem yield
This calculation remains valid with infinite values and avoids subtracting two infinite expectations. In particular, for \(p\geq2\), it gives \(\mathbb{E}A_\infty^p\leq p\mathbb{E}[ZA_\infty^{p-1}]\).
If \(M\) is a continuous \(L^2\)-bounded martingale with \(M_0=0\), take \(A=\langle M\rangle\), \(Z=M_\infty^2\). The stopped square martingale is uniformly integrable: \(M_T^2\leq\mathbb{E}[M_\infty^2\mid\mathcal F_T]\), and \(A_T\leq A_\infty\in L^1\). Optional sampling then gives
If \(\mathbb{E}|M_\infty|^{2p}<\infty\), stop first at \(\rho_n=\inf\{t:A_t\geq n\}\). The stopped bracket has finite \(p\)-th moment, so Hölder’s inequality can be divided through without an infinity assumption:
Conditional Jensen was used in the last step, since \(M_{\rho_n}=\mathbb{E}[M_\infty\mid\mathcal F_{\rho_n}]\). Monotone convergence gives \(\mathbb{E}\langle M\rangle_\infty^p\leq p^p\mathbb{E}|M_\infty|^{2p}\). Thus, for a continuous local martingale \(N_0=0\) and a stopping time \(T\) such that \(N^T\) is uniformly integrable,
If the right side is finite, \(N^T_t=\mathbb{E}[N_T\mid\mathcal F_t]\) is \(L^{2p}\)-bounded and the preceding proof applies; otherwise the bound is automatic. For \(T=\infty\), \(N_T\) denotes the terminal limit of the uniformly integrable stopped martingale.
Uniform integrability is essential. For Brownian motion and \(a\ne0\), \(T_a=\inf\{t:B_t=a\}<\infty\) almost surely, but \(\mathbb{E}T_a=\infty\). Indeed, the reflection principle gives \(\mathbb P(T_a>t)=2\Phi(|a|/\sqrt t)-1\sim\sqrt{2/\pi}|a|/\sqrt t\), whose integral diverges. Hence \(\mathbb{E}\langle B\rangle_{T_a}^p=\infty\) for \(p\geq1\), while \(|B_{T_a}|^{2p}=|a|^{2p}\). The stopped process is not uniformly integrable, since that would imply \(0=\mathbb{E}B_0=\mathbb{E}B_{T_a}=a\).
1.1. Further Developments on Stochastic Integration
Following (Le Gall 2016), for \(L^2\)-bounded martingales \(M\), define the Hilbert space
Clearly, \(M \in \mathcal H^2\) if and only if \(M_0 = 0\) and \(\mathbb{E}\langle M,M\rangle_\infty < \infty\), so \(M_t = \mathbb{E}[M_\infty \mid \mathcal F_t]\) for all \(t\), and \(M_\infty\) is the almost sure limit of \(M_t\). We can define the scalar product
which is a symmetric bilinear form on \(\mathcal H^2\). We can show that \((\mathcal H^2, (\cdot,\cdot)_{\mathcal H^2})\) is a Hilbert space, i.e. every Cauchy sequence in \(\mathcal H^2\) has a limit in \(\mathcal H^2\).
The completeness assertion can be checked at terminal values. If \((M^n)\) is Cauchy in \(\mathcal H^2\), then \(M_\infty^n\) is Cauchy in \(L^2\); write its limit as \(Z\). Set \(M_t=\mathbb E[Z\mid\mathcal F_t]\). Conditional expectation is a contraction, and Doob’s inequality gives
Choose a subsequence with summable expected uniform differences. It converges uniformly almost surely to a continuous adapted process. At each fixed time this agrees with the conditional-expectation limit, so it supplies a continuous version of \(M\). Also \(\mathbb E[Z\mid\mathcal F_0]=0\), and \(\|M^n-M\|_{\mathcal H^2}=\|M_\infty^n-Z\|_2\to0\). Thus \(\mathcal H^2\) is complete.
For the bracket criterion used here, let \(M\) be a zero-start continuous local martingale with \(\mathbb E\langle M\rangle_\infty<\infty\). Choose increasing stops \(\sigma_n\) making \(M^{\sigma_n}\) bounded, with deterministic cutoffs. The bounded-martingale square identity and optional sampling give, for \(m\geq n\),
The integrable bracket dominates the differences. The terminal values therefore have an \(L^2\) limit, and their closed martingales converge as in the preceding paragraph to \(M\). Conversely, if \(M\in\mathcal H^2\), stopping first and using \(\sup_t|M_t|^2\in L^1\) gives \(\mathbb E\langle M\rangle_\infty=\mathbb E M_\infty^2\). Polarization yields the stated scalar-product identity.
In the main notes, we showed that we can define stochastic integrals for elementary processes (defined based on predictable processes \(H\)). For elementary processes,
where \(\mathcal P\) is the predictable \(\sigma\)-field on \(\Omega \times \mathbb R_+\), and \(\dif P \dif \, \langle M, M \rangle_s\) assigns mass \(\mathbb{E}\int_0^\infty \mathbb 1_A ((\omega,s)) \dif \, \langle M, M \rangle_s\) to any \(A \in \mathcal P\). This is equivalent to an upcoming definition of \(L^2 (M)\). A brief recap of the theory of stochastic integration for elementary processes:
The set of all elementary processes (defined based on predictable processes \(H\)) defines a linear subspace of \(L^2 (M)\), denoted \(\mathcal E\), is dense in \(L^2 (M)\).
For a stopped process of \(M \in \mathcal H^2\), \(\langle M^T, M^T \rangle_\infty = \langle M, M \rangle_T\). Furthermore, for \(H \in L^2 (M)\), \((\mathbb 1_{[0,T]} H)_s (\omega) = \mathbb 1_{\{0 \leq s \leq T(\omega)\}} H_s (\omega)\) also belongs to \(L^2 (M)\), because \(\mathbb 1_{[0,T]}\) is a predictable process—it is adapted to \(\mathcal F_t\) and has left-continuous sample paths.
For every \(M \in \mathcal H^2\) and \(H \in \mathcal E\), the formula
\[\begin{aligned} (H \cdot M)_t = \sum_{0 \leq i \leq n-1} H_{(i)} (M_{t_{i+1} \wedge t} - M_{t_i \wedge t}) \end{aligned}\]defines a process \(H \cdot M \in \mathcal H^2\). One can also show that the map \(H \mapsto H \cdot M\) is an isometry from \(L^2 (M) \to \mathcal H^2\), and that \(H \cdot M\) is the unique martingale of \(\mathcal H^2\) such that \(\langle H \cdot M, N \rangle = H \cdot \langle M, N \rangle\) for all \(N \in \mathcal H^2\). The following stopping time property also holds: \((\mathbb 1_{[0,T]} H) \cdot M = (H \cdot M)^T = H \cdot M^T\). Here, \((H \cdot M)_t = \int_0^t H_s \dif M_s\) is the stochastic integral of \(H\) with respect to \(M\).
The Hilbert-space construction gives the integral under a global second-moment bound. Stopping extends it to a continuous local martingale \(M\). To that end, define
With these definitions, we can prove a number of things.
Proposition 9 (Le Gall (2016, Theorem 5.6)). Let \(M\) be a continuous local martingale. For all \(H \in L^2_{\rm loc} (M)\), there exists a unique continuous local martingale starting at zero, denoted \(H \cdot M\), such that, for all continuous local martingale \(N\),
The following stopping time property also holds: \((\mathbb 1_{[0,T]} H) \cdot M = (H \cdot M)^T = H \cdot M^T\). Furthermore, if \(H \in L^2_{\rm loc} (M)\) and \(K\) is a predictable process, then \(K \in L^2_{\rm loc} (H \cdot M)\) if and only if \(HK \in L^2_{\rm loc} (M)\). Then, the following associativity property holds: \(K \cdot (H \cdot M) = (HK) \cdot M\). The stochastic integral defined here agrees with that defined for elementary processes if \(M \in \mathcal H^2\) and \(H \in L^2 (M)\).
Proof of the local integral extension
First let \(M,N\in\mathcal H^2\) and \(H\in L^2(M)\). For a bounded elementary \(H\), the increment-product characterization of covariation gives
For one elementary interval, include its endpoints in the approximating partitions. On that interval the increments of the integral are the increments of \(M\) multiplied by its fixed left-endpoint coefficient. The covariation sums therefore give the identity directly, and finite sums give it for every elementary \(H\).
For general \(H\), choose elementary \(H^n\to H\) in \(L^2(M)\). The isometry gives \(H^n\cdot M\to H\cdot M\) in \(\mathcal H^2\). The Kunita–Watanabe inequality bounds the expected total variations of the two errors by
Both tend to zero, proving the identity. If \(I,J\in\mathcal H^2\) both have the asserted covariations, take \(N=I-J\). Their difference has zero bracket and is zero by Lemma 5. This proves uniqueness.
The stopping identity does not require a stopping time to have deterministic values. For a bounded grid-valued stopping time \(T\), the process \(\mathbb1_{[0,T]}\) is elementary on that grid: its coefficient on \((t_{j-1},t_j]\) is \(\mathbb1_{\{T\geq t_j\}}\in\mathcal F_{t_{j-1}}\). Refining with the elementary intervals of \(H\) proves the identity by finite sums. Round a bounded general \(T\) upwards to successively finer grids. The indicators converge pointwise to \(\mathbb1_{[0,T]}\), while path continuity gives convergence of the stopped integrals. The isometry and dominated convergence in \(\mathbb E\,d\langle M\rangle\) identify the two limits. Approximate a general \(H\) in \(L^2(M)\), and then let bounded time cutoffs increase. This proves
Now center a general local martingale \(M\) by subtracting \(M_0\), which does not affect its increments or bracket. Write \(A=\langle M\rangle\) and \(C_t=\int_0^tH_s^2\,dA_s\). The assumed local integrability makes \(C\) finite and continuous on every bounded interval. The stopping times
increase to infinity. The stopped integrator is in \(\mathcal H^2\) by the bracket criterion, and \(H\) has squared integral norm at most \(n\) with respect to it. Construct \(I^n=H\cdot M^{\tau_n}\) using the global theory. The stopping identity gives \((I^{n+1})^{\tau_n}=I^n\), so these continuous martingales agree on overlaps and define a continuous local martingale \(I\) with \(I^{\tau_n}=I^n\). This construction is independent of the chosen stops, by comparison on common stopped intervals.
For any continuous local martingale \(N\), stop further so that its centered bracket has bounded expectation. The global covariation identity applies to these common stops, and removing them gives \(\langle I,N\rangle=H\cdot\langle M,N\rangle\). Taking \(N\) equal to a difference of two candidates again proves uniqueness. Taking first \(N=M\) and then \(N=I\) gives \(\langle I\rangle=H^2\cdot\langle M\rangle\). Therefore
This proves the equivalence of the two local-integrability conditions. Finally \(\langle K\cdot I,N\rangle=KH\cdot\langle M,N\rangle\) for every local martingale \(N\). Uniqueness gives \(K\cdot(H\cdot M)=(KH)\cdot M\), completing Proposition 9. The construction and estimates above supply the proof of the result attributed to Le Gall, without relying on an unstated extension theorem.
We can now integrate against each part of a continuous semimartingale: use the local-martingale construction for one part and the pathwise finite-variation integral for the other. If \(X\) is a continuous semimartingale, then \(X\) admits the canonical decomposition into \(X = X_0 + M + V\), where \(M\) is a local martingale, \(V\) is a finite variation process, and \(M_0 = V_0 = 0\). For a locally bounded predictable process \(H\), the stochastic integral \(H \cdot X\) is the continuous semimartingale defined through \(H \cdot X = H \cdot M + H \cdot V\), and
This construction enjoys a number of properties:
Lemma 10. Let \(X\) be a continuous semimartingale, and \(H\) be a locally bounded predictable process. The stochastic integral of \(H\) with respect to \(X\), as defined, satisfies
\((H,X) \mapsto H \cdot X\) is bilinear.
\(H \cdot (K \cdot X) = (HK) \cdot X\) if \(H\) and \(K\) are locally bounded and predictable processes.
The stopping time property holds: for every stopping time \(T\), \((H \cdot X)^T = H \mathbb 1_{[0,T]} \cdot X = H \cdot X^T\).
If \(X\) is a continuous local martingale, then so is \(H\cdot X\). If \(X\) has finite variation, then \(H\cdot X\) has finite variation: on each bounded interval its total variation is at most \(\int |H_s|\,|dX_s|\), which is finite because \(H\) is locally bounded. This second assertion is a pathwise integration fact.
Let \(0 = t_0 < t_1 < \dots < t_n\), and define the elementary process \(H\) via
\[\begin{aligned} H_s (\omega) = \sum_{0 \leq i \leq n-1} H_{(i)} (\omega) \mathbb 1_{(t_i,t_{i+1}]} (s) ~. \end{aligned}\]Suppose \(H_{(i)}\) is \(\mathcal F_{t_i}\)-measurable for all \(0 \leq i \leq n-1\). Then,
\[\begin{aligned} (H \cdot X)_t = \sum_{0 \leq i \leq n-1} H_{(i)} (X_{t_{i+1} \wedge t} - X_{t_i \wedge t}) ~. \end{aligned}\]
Proof. For elementary bounded integrands, bilinearity and the stopping property follow directly from finite sums. Extension by the isometry proves them for square-integrable integrands, and stopping proves the local versions. The finite-variation parts follow from ordinary Stieltjes integration. Associativity can also be checked through covariation: \(\langle K\cdot(H\cdot M),N\rangle=KH\cdot\langle M,N\rangle\), and uniqueness identifies the integral. Its integrability condition is exactly \(\int K^2\,d\langle H\cdot M\rangle=\int(KH)^2\,d\langle M\rangle\).
For (5), let there be \(m\) elementary intervals, and assume each finite coefficient \(H_{(i)}\) is \(\mathcal F_{t_i}\)-measurable. Define \(\tau_n=\min\{t_i:|H_{(i)}|>n\}\), with minimum of the empty set equal to infinity. This is a stopping time because \(\{\tau_n\leq t\}\) is a finite union of events in \(\mathcal F_t\). Since there are only finitely many coefficients, \(\tau_n\uparrow\infty\) pathwise. Furthermore,
whose coefficients are bounded by \(n\). The bounded elementary formula, combined with a reducing sequence for the integrator’s local-martingale part, therefore applies. Remove both stops to obtain the finite-sum formula. No continuity of the elementary process is asserted or needed. ◻
Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.