Lebesgue-Stieljes Integral

Deterministic Definition

Theorem: Lebesgue-Stieljes measure

Let $F:\mathbb{R}\to \mathbb{R}$ be a càd increasing function function (in particular càdlàg). There exists a unique measure $dF$ on the Borel $\sigma$-algebra of the real line such that

\[ dF((a,b]) = F(b) - F(a) \]

for any real $a\leq b$. This measure is called the Lebesgue-Stieljes measure.

Proof

Let $\Omega = \mathbb{R}$ and $\mathcal{B}$ be the Borel $\sigma$-algebra which is generated by the semi-ring $\mathcal{R} = \{(a,b] \colon a \leq b\}$ with the convention that $(a,a] = \emptyset$. Define

\[ P((a,b]) = F(a) - F(b) \]

for any $(a, b]$ in $\mathcal{R}$. Straightforward inspection shows that $P$ is additive, such that $P[\emptyset] = 0$ and $P$ is sub-additive. To show that $P$ extends uniquely to a $\sigma$-finite measure on $\mathcal{B}$, we just have to check that $P$ is $\sigma$-subadditive on $\mathcal{R}$. Let $A = (a,b]$ be in $\mathcal{R}$ and $(A_n) = (]a_n,b_n])$ a countable family in $\mathcal{R}$ such that $A \subseteq \cup A_n$. Taking $\varepsilon > 0$, by right-continuity of $F$, choose some $a^\varepsilon \in (a,b[$ such that $F(a^\varepsilon) - F(a) < \varepsilon/2$. Also using the right-continuity of $F$, choose $b_n^\varepsilon > b_n$ for every $n$ such that $F(b_n^\varepsilon) - F(b_n) \leq \varepsilon 2^{-n-1}$. It follows that

\[ [a^\varepsilon, b] \subseteq (a,b] \subseteq \cup (a_n, b_n] \subseteq \cup (a_n, b_n^\varepsilon). \]

However, $[a^\varepsilon, b]$ is a compact set, therefore, the open covering $\cup (a_n, b_n^\varepsilon)$ of $[a^\varepsilon, b]$ can be chosen finite. Hence, there exists $n_0$ such that

\[ [a^\varepsilon, b] \subseteq (a^\varepsilon, b] \subseteq \cup_{k \leq n_0} (a_k, b_k^\varepsilon) \subseteq \cup_{k \leq n_0} (a_k, b_k^\varepsilon]. \]

and therefore

\[ \begin{align*} P((a,b]) & = F(b) - F(a)\\ & \leq \varepsilon/2 + F(b) - F(a^\varepsilon)\\ & \leq \varepsilon/2 + \sum_{k=1}^{n_0} (F(b_k^\varepsilon) - F(a_k)) \\ & \leq \varepsilon + \sum (F(b_n) - F(a_n))\\ & = \varepsilon + \sum P((a_n, b_n]). \end{align*} \]

showing that $P$ extends to a measure on the real line. This measure is also $\sigma$-finite in the sense that there exists an increasing sequence of sets $(]a_n,b_n])$ such that $\mathbb{R} = \cup ]a_n, b_n]$ and $P(]a_n, b_n]) < \infty$ for every $n$. Hence, this extension is unique and we denote it $dF$.

Example

The classical Lebesgue measure $dx$ on the real line is derived from the continuous increasing function $F(x) = x$, for which it holds

\[ dx((a,b]) = b - a \]
If we consider the increasing right continuous function $F(x) = 1_{[y,\infty)}(x)$ for a given real $y$ we get the Dirac measure

\[ dF[A] = \delta_y[A] = \begin{cases} 1 & \text{if } y \in A \\ 0 & \text{otherwise} \end{cases} \]

Clearly, given two càd increasing function, their difference defines a signed measure. The set of those functions are better described as the set of bounded variation functions.

Definition

We say that a function $F:[0,\infty) \to \mathbb{R}$, $t \mapsto F(t) := F_t$ is of bounded variation if

\[ S_t = \sup_{\Pi \text{ subdivision of } [0,t]} S_t^{\Pi} < \infty \]

for every $t$ where for $\Pi = \{0=t_0<t_1<\ldots<t_n = t\}$

\[ S_t^{\Pi} = \sum_{1 \leq k \leq n} |F_{t_{k+1}} - F_{t_k}| \]

Functions of bounded variations are those that can be defined as a difference of two increasing functions.

Proposition

A function is of bounded variations if and only if it can be written as the difference between two increasing functions.

Proof

Let $F$ be a function of bounded variations. Inspection shows that $F^+ = (S + F)/2$ and $F^- = (S - F)/2$ are two increasing functions whose difference is equal to $F$. This decomposition is actually the minimal one, in the sense that if $F = A - B$ for two increasing functions $A$ and $B$, then it follows that $F^+ \leq A$ and $F^- \leq B$. The reciprocal is easy.

Given a right continuous function $F$ of bounded variations, we can therefore define a so-called signed measure and the absolute value of this measure

\[ dF = dF^+ - dF^- \quad \text{and} \quad |dF| = dF^+ + dF^- \]

If $H:[0,\infty) \to \mathbb{R}$ is a locally bounded, that is, bounded on any compact interval $[0,t]$, and $\mathcal{B}([0,\infty))$-measurable, we can define the integral

\[ \int_{0}^{t} H_s dF_s = \int_{(0,t]} H_s dF_s := \int_{0}^{t} H_s dF^+_s - \int_{0}^{t} H_s dF^-_s \]

which is called the Lebegues-Stieljes integral of $H$ with respect to $F$. The integral is understood over the interval $]0,t]$ so that $\int_0^t dF_s = F_t - F_0$.

Proposition: Chain Rule or Integration by Parts

Let $F$ and $G$ be two right continuous functions of finite variations, then it holds

\[ \begin{align*} F_t G_t & = F_0 G_0 + \int_{0}^{t} F_s dG_s + \int_{0}^{t} G_{s-} dF_s\\ & = F_0 G_0 + \int_{0}^{t} F_{s-} dG_s + \int_{0}^{t} G_{s} dF_s\\ & = F_0 G_0 + \int_{0}^{t} F_{s-} dG_s + \int_{0}^{t} G_{s-} dF_s + \sum_{0<s \leq t} \Delta F_s \Delta G_s \end{align*} \]

where $F_{s-} = \lim_{u \nearrow s} F_u$ and $\Delta F_s = F_s - F_{s-} = dF[\{s\}]$.

Remark

This proposition is often stated in differential form, that is

\[\begin{equation*} dFG = F_{\cdot -}dG + G_{\cdot -}dF + \Delta F \Delta G \end{equation*}\]

Note that if $F$ and $G$ are continuous, the more classical chain rule formula reads as follows

\[\begin{equation*} dFG = FdG + G dF \end{equation*}\]

Proof

Considering the product measure $dF \otimes dG$ on $[0,\infty) \times [0,\infty)$, using the triangular equality

\[ 1_{]0,t]}(s_1) 1_{]0,t]}(s_2) = 1_{]0,t]}(s_1) 1_{]0,s_1]}(s_2) + 1_{]0,s_2[}(s_1) 1_{]0,t]}(s_2) \]

using Fubini-Tonelli, we obtain

\[ \begin{align*} dF \otimes dG \left[ (0,t] \times (0,t] \right] & = (F_t - F_0)(G_t - G_0) \\ & = \int \int 1_{]0,t]}(s_1) 1_{]0,t]}(s_2) \, dF_{s_1} dG_{s_2} \\ & = \int_{(0,t]} \left( \int_{(0,s_1]} dG_{s_2} \right) dF_{s_1} + \int_{(0,t]} \left( \int_{(0,s_2)} dF_{s_1} \right) dG_{s_2} \\ & = \int_{(0,t]} \left(G_s - G_0\right) \, dF_s + \int_{(0,t]} \left(F_{s-} - F_0\right) \, dG_s \\ & = \int_{(0,t]} G_s \, dF_s + \int_{(0,t]} F_{s-} \, dG_s - G_0 (F_t - F_0) - F_0 (G_t - G_0) \end{align*} \]

showing the first equality. The second follows exactly the same argumentation by swapping $F$ and $G$. As for the last one, note that $F = F_{-} + \Delta F$, and since $F$ can only have countably many discontinuity points, the last equality follows.

This basic form of the classical chain rule formula leads to the general one

Theorem: Chain Rule Formula

Let $f:\mathbb{R} \to \mathbb{R}$ be a continuously differentiable function and $F\colon [0, \infty)\to \mathbb{R}$ be a càd bounded variation function. It follows that

\[ f(F_t) = f(F_0) + \int_{0}^{t} f^\prime(F_{s-}) \, dF_s + \sum_{s \leq t} \left( f(F_s) - f(F_{s-}) - f^\prime(F_{s-}) \Delta F_s \right) \]

In particular, if $F$ is continuous, it holds

\[ f(F_t) = f(F_0) + \int_{0}^{t} f^\prime(F_s) \, dF_s \]

Remark

The differential form of this chain rule formula reads as follows

\[ \begin{equation*} df(F) = f^\prime(F_{\cdot -}) dF + \Delta f(F) - f^\prime(F_{\cdot -})\Delta F \end{equation*} \]

and if $F$ is continuous we obtain the classical chain rule formula

\[ df(F) = f(F)dF \]

which is the building block for differential equations.

Proof

We here just sketch the idea of the proof, as a more general one will be shown for the Ito-Formula later in the lecture.

Step 1: Any differentiable function $ f \colon \mathbb{R} \to \mathbb{R}$ can be approximated on any compact uniformly by interpolation by a polynomial $f(x) = \sum_{k\leq n} \alpha_k x^k$. Hence, it would be enough to show it for polynomial of arbitrary degree.
Step 2: Since the integral is a linear operator, showing the formula for any polynomial is equivalent to show it for any monomial $x^n$. We show by induction that the chain rule formula holds for any monomial $x^n$, that is

\[ \begin{equation*} d F^n = n F^{n-1}_{\cdot -} dF + \Delta F^n - n F^{n-1}_{\cdot -}\Delta F \end{equation*} \]

For $n=1$, this is immediate. Suppose that it holds for any $m \leq n-1$ and we show if for $n\geq 2$. Defining $G = F^{n-1}$, from the product chain rule formula and the recursion hypothesis for $G$ it holds that

\[ \begin{align*} dF^n & = d FG \\ & = F_{\cdot -} dG + G_{\cdot -} dF + \Delta F \Delta G\\ & = F_{\cdot -}\left( (n-1) F^{n-2}_{\cdot -} dF + \Delta F^{n-1} - (n-1)F^{n-2}_{\cdot -}\Delta F \right) + F^{n-1}_{\cdot -}dF + \Delta F \Delta F^{n-1}\\ & = n F^{n-1}dF + F_{\cdot -}F^{n-1} - F_{\cdot - }^n -(n-1)F^{n-1}_{\cdot -}F + (n-1)F^n_{\cdot -}\\ & \quad \quad \quad \quad + F^n + F^n_{\cdot -} - F_{\cdot -}F^{n-1} - F F^{n-1}_{\cdot -}\\ & = n F^{n-1}dF + \Delta F^n - n F^{n-1}_{\cdot -} \Delta F \end{align*} \]

Remark

Note that according to the proof, we can show the chain rule formula for any monomial of the type $x^n y^m$. Hence the chain rule formula extends for multifariate functions $f$ which in the case of continuous bounded càd functions $F$ and $G$ yields

\[\begin{equation*} df(F, G) = \partial_x f(F, G)dF +\partial_y f(F, G) \end{equation*}\]

Stochastic Lebesgue-Stieljes Integral

We can extend this integration procedure for every $\omega$-dependent paths as follows.

Definition

A process $A$ is called increasing if
- $A$ is adapted with $A_0 = 0$
- $A$ is càdlàg, and almost all sample paths are increasing

An increasing process $A$ is called integrable if $E[A_t] < \infty$.

A process $A$ is called of bounded variations if $A$ is the difference between two increasing processes.

We denote by $dA$ the $\omega$-wise $\sigma$-finite signed measure $dA_t(\omega)$ induced by $A$. For every locally measurable process(1) process $H$ which is locally bounded, that is $(\omega,s) \mapsto H_s(\omega)$ is uniformly bounded for almost all $\omega$ and all $s \in [0,t]$, we can define the $\omega$-wise Lebesgue-Stieljes Integral

Recall that a measurable process is a process such that $(\omega,t) \mapsto H_t(\omega)$ is $\mathcal{F} \otimes \mathcal{B}([0,\infty))$-measurable; in particular $t \mapsto H_t(\omega)$ is $\mathcal{B}([0,\infty[)$-measurable for every $\omega$.

\[ \int_0^t H_s(\omega) \, dA_s(\omega) \]

for any $\omega$ and $0\leq t<\infty$. We usually do not mention the time and omega index and simplify the notation to $\int_{0}^{t} H \, dA$.

Proposition

If $H$ is a locally bounded measurable process and $A$ is a process of bounded variations, then

\[ \int H \, dA = \left( \int_{0}^{t} H \, dA \right) \]

defines a càdlàg right continuous measurable process. If furthermore $H$ is progressive, then $\int H \, dA$ is progressive, càdlàg, and of bounded variations.

Proof

The proof is quite easy, as one approximates $H$ by sequences of simple step processes with the right measureability. Let us however check that if $H$ is progressive, then $\int H \, dA$ is of bounded variations. Since it is càdlàg and adapted with $A_0 = 0$, we just have to check that it is of bounded variations. The process $H$ being locally bounded, let $K$ be such that $|H_s(\omega)| \leq K$ for every $0\leq s\leq t$. For any partition $\Pi$ of $[0,t]$, it holds

\[ \begin{align*} \sum_{\Pi} \left| \int_{t_{k-1}}^{t_k} H_s(\omega) \, dA_s(\omega) \right| & \leq \sum_{\Pi} \int_{t_{k-1}}^{t_k} |H_s(\omega)| \, d|A_s(\omega)|\\ & \leq K \sum_{\Pi} \int_{t_{k-1}}^{t_k} d|A_s(\omega)| \\ & = K |A_t(\omega)|< \infty \end{align*} \]

showing that $\int X \, dA$ is of bounded variations.

Exercice

Using Radon-Nikodym, show that for any two increasing processes $A$ and $B$ such that $A - B$ is still increasing, then there exists an adapted measurable process $H$ such that

\[ B = \int H dA \]

In particular, if $A$ is of bounded variations, there exists $H$ adapted and measurable such that

\[ A = \int H \left| dA \right| \]