Integrability Criteria for Continuous Semimartingales

Two Parts, Two Integrability Tests

For a continuous semimartingale, an integrand must be compatible with two different sources of variation. The local martingale part is measured by its quadratic variation; the finite-variation part is measured by total variation. Checking only one of these conditions can give the wrong answer, even when the other condition is satisfied.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$ carrying a Brownian motion $B_t$. Let $X=M+A$ be a continuous semimartingale decomposition, where $M$ is a continuous local martingale and $A$ is a continuous adapted process of finite variation on every bounded time interval. Write $[M]_t$ for the quadratic variation of $M$, and $V_t(A)$ for the total variation of $A$ on $[0,t]$. The criterion below applies to a predictable integrand $H$; in particular, a continuous adapted $H$ is predictable. Deterministic Borel functions of time used in the examples are predictable.

Definition: The candidate $H$ is locally integrable against $X=M+A$ if, almost surely, for every finite $T$, $$ \int_0^T H_s^2\,d[M]_s<\infty \qquad\text{and}\qquad \int_0^T |H_s|\,dV_s(A)<\infty. $$ These are pathwise Lebesgue–Stieltjes integrals against increasing processes. The first condition controls the local martingale part; the second controls the finite-variation part.

“Almost surely finite” is not the same as “has finite expectation.” This criterion does not require $\mathbb{E}\int_0^T H_s^2\,d[M]_s<\infty$. For comparison, the usual Brownian Itô integral $\int_0^T H_s\,dB_s$ is defined on the class of progressively measurable $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for an $L^2$ integral. The semimartingale criterion is a pathwise local condition and can apply even when an expectation is infinite.

The Integrability Criterion

Standard result (continuous-semimartingale integrability criterion): For a continuous semimartingale $X=M+A$ and a predictable process $H$, $H$ is locally integrable against $X$ precisely when, almost surely on every bounded time interval, $$ \int_0^T H_s^2\,d[M]_s+\int_0^T |H_s|\,dV_s(A)<\infty. $$ The integral with respect to $X$ is then constructed by combining the local-martingale integral with the pathwise finite-variation integral. The construction proof belongs to the subsequent development of stochastic integration against continuous semimartingales.

The two summands in this test cannot be substituted for one another. The bracket integral is the condition for the local-martingale component; the variation integral is the condition for the finite-variation component. In particular, the drift part does not become harmless merely because the bracket is small or zero. Nor does a finite variation integral compensate for an infinite bracket integral.

It is useful to collect the two tests into one increasing process, $$ C_t(H;M,A)=\int_0^t H_s^2\,d[M]_s+\int_0^t |H_s|\,dV_s(A). $$ Then the criterion is that $C_T(H;M,A)<\infty$ almost surely for every finite $T$. Since $C$ is increasing, it is enough to check this for positive integer $T$: finiteness on $[0,n]$ implies finiteness on every $[0,T]$ with $T\le n$.

Proposition: Suppose $C_t=C_t(H;M,A)$ has continuous increasing paths. Almost-sure finiteness of $C_T$ for every finite $T$ is equivalent to the existence of stopping times $\tau_n$ increasing to infinity almost surely such that $C_{t\wedge\tau_n}\le n$ for every $t$, after a harmless adjustment of the index if needed.

Proof. First suppose $C_T<\infty$ almost surely for every finite $T$. Define $\tau_n=\inf\{t\ge0:C_t\ge n\}$, with the infimum of the empty set taken to be infinity. Continuity and monotonicity imply $C_{t\wedge\tau_n}\le n$. For each fixed $T$, the finite random variable $C_T$ is eventually less than $n$, so $\tau_n>T$ for all sufficiently large $n$. Taking integer $T$ shows $\tau_n\to\infty$ almost surely. Conversely, suppose such stopping times exist and $\tau_n\to\infty$ almost surely. Fix a finite $T$. On the event $\{\tau_n\ge T\}$, $C_T=C_{T\wedge\tau_n}\le n$. Almost surely, some $n$ has $\tau_n\ge T$, so $C_T$ is finite. This holds for every integer $T$, and monotonicity gives the result for every finite $T$. $\square$

This stopping-time formulation is useful because local conditions need not give one deterministic bound that works on every path. Stopping instead restricts attention to successive intervals on which the accumulated integrability test remains bounded.

Proposition: If $|H_t|\le K$ almost surely for all $t\le T$, where $K$ is a finite deterministic constant, then $H$ satisfies both integrability tests on $[0,T]$, provided $[M]_T$ and $V_T(A)$ are finite almost surely.

Proof. By monotonicity of integration against the nonnegative measures $d[M]$ and $dV(A)$, $$ \int_0^T H_s^2\,d[M]_s\le K^2[M]_T,\qquad \int_0^T |H_s|\,dV_s(A)\le K V_T(A). $$ Both right-hand sides are finite almost surely by assumption. Thus both required integrals are finite almost surely. $\square$

For a continuous local martingale, its quadratic variation is finite on bounded intervals almost surely; for a finite-variation process, its total variation is finite there by definition. Therefore a bounded predictable integrand is a common sufficient case. Boundedness is not necessary: an unbounded integrand can still pass both tests if its growth is sufficiently small relative to the two measures.

Worked Applications

Worked Example: A Bounded Integrand Against Brownian Motion with Drift

Let $X_t=B_t+2t$ on $[0,1]$ and $H_t=1+t$. Identify the two measures and compute both integrability tests.

Here $M_t=B_t$, so $[M]_t=t$ and $d[M]_t=dt$. Also $A_t=2t$, so $V_t(A)=2t$ and $dV_t(A)=2\,dt$. Therefore $$ \int_0^1 H_t^2\,d[M]_t =\int_0^1(1+t)^2\,dt =\left[t+t^2+\frac{t^3}{3}\right]_0^1 =\frac{7}{3}. $$ For the finite-variation part, $$ \int_0^1 |H_t|\,dV_t(A) =2\int_0^1(1+t)\,dt =2\left[t+\frac{t^2}{2}\right]_0^1 =3. $$ Both are finite, so the criterion is satisfied. Check: $H_t$ lies between $1$ and $2$ on this interval, and the computed bounds are consistent with integrating a bounded function against measures of total masses $1$ and $2$, respectively.

Worked Example: A Singular Integrand Against Brownian Motion

Take $X_t=B_t$ on $[0,1]$, and define $H_0=0$ and $H_t=t^{-1/2}$ for $t>0$. Decide whether $H$ meets the criterion. Compare it with $G_0=0$, $G_t=t^{-1/4}$ for $t>0$.

For $B$, $d[M]_t=dt$, and the finite-variation part is zero. For $H$, the bracket test is $$ \int_0^1 H_t^2\,dt=\int_0^1\frac{1}{t}\,dt=\infty, $$ because $\int_\varepsilon^1 t^{-1}\,dt=-\log\varepsilon\to\infty$ as $\varepsilon\downarrow0$. Thus $H$ is not locally integrable against $B$ on $[0,1]$. For $G$, the test gives $$ \int_0^1 G_t^2\,dt=\int_0^1 t^{-1/2}\,dt =\left[2\sqrt{t}\right]_0^1=2. $$ The finite-variation test is zero for both integrands, so $G$ passes and $H$ fails. Check: cutting off the singularity at $\varepsilon>0$ gives $\int_\varepsilon^1 H_t^2\,dt=-\log\varepsilon$, which increases without bound, whereas $\int_0^\varepsilon G_t^2\,dt=2\sqrt{\varepsilon}\to0$.

Worked Example: Both Tests for a Singular Drift

On $[0,1]$, let $M_t=B_t$, $A_t=2\sqrt{t}$, $X_t=M_t+A_t$, and $H_0=0$, $H_t=t^{-1/3}$ for $t>0$. Check both integrability conditions.

The function $A$ is increasing and has $dV_t(A)=dA_t=t^{-1/2}\,dt$ for $t>0$. Its total variation on $[0,1]$ is $2$, which is finite. Since $d[M]_t=dt$, $$ \int_0^1 H_t^2\,d[M]_t =\int_0^1 t^{-2/3}\,dt =\left[3t^{1/3}\right]_0^1=3. $$ The variation test is $$ \int_0^1 |H_t|\,dV_t(A) =\int_0^1 t^{-1/3}t^{-1/2}\,dt =\int_0^1 t^{-5/6}\,dt =\left[6t^{1/6}\right]_0^1=6. $$ Both are finite, so $H$ passes. Check: the contributions from $[0,\varepsilon]$ are $3\varepsilon^{1/3}$ and $6\varepsilon^{1/6}$; each tends to zero as $\varepsilon\downarrow0$, confirming that neither integral diverges at the origin.

Worked Example: A Finite-Variation Test Can Fail by Itself

Let $X_t=t$ on $[0,1]$, so $M=0$ and $A_t=t$. Set $H_0=0$ and $H_t=t^{-1}$ for $t>0$. Does $H$ pass the criterion?

The bracket term is zero because $[M]_t=0$: $$ \int_0^1 H_t^2\,d[M]_t=0. $$ But $dV_t(A)=dt$, and $$ \int_0^1 |H_t|\,dV_t(A) =\int_0^1\frac{1}{t}\,dt =\infty. $$ Thus $H$ fails the criterion. Check: on $[\varepsilon,1]$ the variation integral equals $-\log\varepsilon$, which diverges as the lower endpoint decreases to zero.

A Common Mistake

A common error is to test only the local-martingale component and conclude that a zero bracket integral is enough. In the last example, the incorrect calculation would be: “$M=0$, so $\int_0^1 H_t^2\,d[M]_t=0<\infty$; therefore $H$ is integrable against $X$.” The conclusion is wrong. The finite-variation component contributes the separate term $\int_0^1|H_t|\,dV_t(A)=\infty$, so the full criterion fails.

The reverse issue also matters: a finite variation test alone does not settle integrability against a martingale component. In the Brownian example, the finite-variation term vanishes, but the bracket integral for $H_t=t^{-1/2}$ is infinite. Always identify the decomposition first, then evaluate both terms.

These conditions also clarify the role of expectations. A pathwise finite bracket integral need not have finite expectation, so the criterion for local semimartingale integrability should not be silently replaced by an $L^2$ requirement. Conversely, a finite expectation for one part cannot remove divergence in the other. The distinction is between the pathwise local-integrability test and stronger moment assumptions used for square-integrable martingale integrals.

Check Your Understanding

For each question, keep the bracket and variation tests separate, and specify whether the condition is pathwise or in expectation.

If $X_t=B_t+3t$ on $[0,1]$ and $H_t=2$, compute both integrability terms.
For $X=B$ on $[0,1]$, test $H_t=t^{-\alpha}$ for $t>0$, with $H_0=0$. For which $\alpha$ is the bracket integral finite?
For $X_t=t$ and $H_t=t^{-1/2}$ on $(0,1]$, compute the variation integral and decide whether the criterion holds.
Explain why finiteness of $\int_0^T H_s^2\,d[M]_s$ does not, by itself, imply integrability against $X=M+A$.
State how the stopping times $\tau_n=\inf\{t:C_t\ge n\}$ localize the increasing integrability process $C_t$.