# Almost Sure

## 8 November 16

### Measurable Projection and the Debut Theorem

I will discuss some of the immediate consequences of the following deceptively simple looking result.

Theorem 1 (Measurable Projection) If ${(\Omega,\mathcal{F},{\mathbb P})}$ is a complete probability space and ${A\in\mathcal{B}({\mathbb R})\otimes\mathcal{F}}$ then ${\pi_\Omega(A)\in\mathcal{F}}$.

The notation ${\pi_B}$ is used to denote the projection from the cartesian product ${A\times B}$ of sets A and B onto B. That is, ${\pi_B((a,b)) = b}$. As is standard, ${\mathcal{B}({\mathbb R})}$ is the Borel sigma-algebra on the reals, and ${\mathcal{A}\otimes\mathcal{B}}$ denotes the product of sigma-algebras.

Theorem 1 seems almost obvious. Projection is a very simple map and we may well expect the projection of, say, a Borel subset of ${{\mathbb R}^2}$ onto ${{\mathbb R}}$ to be Borel. In order to formalise this, we could start by noting that sets of the form ${A\times B}$ for Borel A and B have an easily described, and measurable, projection, and the Borel sigma-algebra is the closure of the collection such sets under countable unions and under intersections of decreasing sequences of sets. Furthermore, the projection operator commutes with taking the union of sequences of sets. Unfortunately, this method of proof falls down when looking at the limit of decreasing sequences of sets, which does not commute with projection. For example, the decreasing sequence of sets ${S_n=(0,1/n)\times{\mathbb R}\subseteq{\mathbb R}^2}$ all project onto the whole of ${{\mathbb R}}$, but their limit is empty and has empty projection.

There is an interesting history behind Theorem 1, as mentioned by Gerald Edgar on MathOverflow (1) in answer to The most interesting mathematics mistake? In a 1905 paper, Henri Lebesgue asserted that the projection of a Borel subset of the plane onto the line is again a Borel set (Lebesgue, (3), pp 191–192). This was based on the erroneous assumption that projection commutes with the limit of a decreasing sequence of sets. The mistake was spotted, in 1916, by Mikhail Suslin, and led to his investigation of analytic sets and to begin the study of what is now known as descriptive set theory. See Kanamori, (2), for more details. In fact, as was shown by Suslin, projections of Borel sets need not be Borel. So, by considering the case where ${\Omega={\mathbb R}}$ and ${\mathcal{F}=\mathcal{B}({\mathbb R})}$, Theorem 1 is false if the completeness assumption is dropped. I will give a proof of Theorem 1 but, as it is a bit involved, this is left for a later post.

For now, I will state some consequences of the measurable projection theorem which are important to the theory of continuous-time stochastic processes, starting with the following. Throughout this post, the underlying probability space ${(\Omega,\mathcal{F})}$ is assumed to be complete, and stochastic processes are taken to be real-valued, or take values in the extended reals ${\bar{\mathbb R}={\mathbb R}\cup\{\pm\infty\}}$, with time index ranging over ${{\mathbb R}_+}$. For a first application of measurable projection, it allows us to show that the supremum of a jointly measurable processes is measurable.

Lemma 2 If X is a jointly measurable process and ${S\in\mathcal{B}(\mathbb{R}_+)}$ then ${\sup_{s\in S}X_s}$ is measurable.

Proof: Setting ${U=\sup_{s\in S}X_s}$ then, for each real K, ${U > K}$ if and only if ${X_s > K}$ for some ${s\in S}$. Hence,

$\displaystyle U^{-1}\left((K,\infty]\right)=\pi_\Omega\left((S\times\Omega)\cap X^{-1}\left((K,\infty]\right)\right).$

By the measurable projection theorem, this is in ${\mathcal{F}}$ and, as sets of the form ${(K,\infty]}$ generate the Borel sigma-algebra on ${\mathbb{\bar R}}$, U is ${\mathcal{F}}$-measurable. ⬜

Next, the running maximum of a jointly measurable process is again jointly measurable.

Lemma 3 If X is a jointly measurable process then ${X^*_t\equiv\sup_{s\le t}X_s}$ is also jointly measurable.

Proof: The process ${Y_t\equiv\sup_{s < t}X_s}$ is left-continuous, and is measurable by Lemma 2. So, Y is jointly measurable. Then, ${X^*=X\vee Y}$ is jointly measurable. ⬜

Lemma 2 extends in the obvious way to progressively measurable processes. For the remainder of this post, we work with respect to a complete filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\in{\mathbb R}_+},{\mathbb P})}$.

Lemma 4 If X is a progressively measurable process and ${S\in\mathcal{B}([0,t])}$ for some time t, then ${\sup_{s\in S}X_s}$ is ${\mathcal{F}_t}$-measurable.

Proof: By definition of progressive measurability, for each time t, ${1_{[0,t]}X}$ is ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$-measurable. Hence, by Lemma 2,

$\displaystyle \sup_{s\in S}X_s=\sup_{s\in S}1_{\{s\in[0,t]\}}X_s$

is ${\mathcal{F}_t}$-measurable. ⬜

We can also show that the running maximum of a progressively measurable process is itself progressive.

Lemma 5 If X is an a progressively measurable process then ${X^*_t\equiv\sup_{s\le t}X_s}$ is itself progressively measurable so, in particular, is adapted.

Proof: By definition, for any fixed time ${t\ge0}$, the process ${1_{[0,t]}X}$ is ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$-measurable. Lemma 3 then says that ${1_{[0,t]}X^*}$ is also ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$-measurable, so ${X^*}$ is progressive. ⬜

It is often useful to look at the left or right limits of stochastic processes. As arbitrary processes do not necessarily have paths for which these limits exist, we look at the limit supremum. The limit supremum and left/strict-left/right/strict-right limits of a process are defined as

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\limsup_{s\rightarrow t}X_s&\displaystyle=\lim_{\epsilon\rightarrow0}\sup_{s\in(t-\epsilon,t+\epsilon)}X_s,\smallskip\\ \displaystyle\limsup_{s\uparrow t}X_s&\displaystyle=\lim_{\epsilon\rightarrow0}\sup_{s\in(t-\epsilon,t]}X_s,\smallskip\\ \displaystyle\limsup_{s\uparrow\uparrow t}X_s&\displaystyle=\lim_{\epsilon\rightarrow0}\sup_{s\in(t-\epsilon,t)}X_s,\smallskip\\ \displaystyle\limsup_{s\downarrow t}X_s&\displaystyle=\lim_{\epsilon\rightarrow0}\sup_{s\in[t,t+\epsilon)}X_s,\smallskip\\ \displaystyle\limsup_{s\downarrow\downarrow t}X_s&\displaystyle=\lim_{\epsilon\rightarrow0}\sup_{s\in(t,t+\epsilon)}X_s. \end{array}$

To be clear, when ${t=0}$ I take ${X_s=X_0}$ for ${s < 0}$ in these definitions so that, for example, ${\limsup_{s\uparrow\uparrow0}X_s=X_0}$. Measurable projection enables us to show that these processes are progressively measurable processes whenever X is.

Lemma 6 If X is a progressively measurable process then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle t&\displaystyle\mapsto\limsup_{s\uparrow t}X_s,\smallskip\\ \displaystyle t&\displaystyle\mapsto\limsup_{s\uparrow\uparrow t}X_s \end{array}$

are progressively measurable. If, furthermore, the filtration ${\mathcal{F}_\cdot}$ is right-continuous, then

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle t&\displaystyle\mapsto\limsup_{s\downarrow t}X_s,\smallskip\\ \displaystyle t&\displaystyle\mapsto\limsup_{s\downarrow\downarrow t}X_s,\smallskip\\ \displaystyle t&\displaystyle\mapsto\limsup_{s\rightarrow t}X_s \end{array}$

are progressively measurable.

Proof: For each positive integer n, choose a sequence of times ${0=t^n_0 < t^n_1 < \cdots}$ increasing to infinity, and such that the mesh ${\sup_k(t^n_k-t^n_{k-1})}$ goes to zero as n goes to infinity. For example, ${t^n_k=k/n}$. Define the process ${Y^n}$ by ${Y^n_0=X_0}$ and

$\displaystyle Y^n_t=\sup\left\{X_s\colon s\in[t^n_{k-1},t)\right\}.$

for t in the interval ${(t^n_{k-1},t^n_k]}$. This is left-continuous and, by Lemma 4, is adapted. So, it is progressive and, hence,

$\displaystyle \limsup_{s\uparrow\uparrow t}X_s=\lim_{n\rightarrow\infty}Y^n_t$

is progressive. This also implies that

$\displaystyle \limsup_{s\uparrow t}X_s = X_t\vee\limsup_{s\uparrow\uparrow t}X_s$

is progressive. Now, with times ${t^n_k}$ as above, define the processes ${Z^n}$ by

$\displaystyle Z^n_t=\sup\left\{X_s\colon s\in(t,t^n_k]\right\}$

for t in the interval ${[t^n_{k-1},t^n_k)}$. This is right-continuous and ${Z_t\equiv\lim_{n\rightarrow\infty}Z^n_t}$ is equal to ${\limsup_{s\downarrow\downarrow t}X_s}$. By Lemma 4, ${X_t}$ is ${\mathcal{F}_{t_k}}$-measurable over ${t < t^n_k}$. Choosing any times ${t > s \ge 0}$, for all large enough n there exists k with ${t\ge t^n_k > s}$. So, ${1_{[0,s]}Z^n}$ will be ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$-measurable. Letting n go to infinity and then letting s increase to t shows that ${1_{[0,t)}Z}$ is ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$-measurable. Furthermore, ${Z_t}$ is ${\mathcal{F}_{t+\epsilon}}$-measurable for all positive ${\epsilon}$, so is ${\mathcal{F}_{t+}}$-measurable. Then,

$\displaystyle 1_{[0,t]}Z=1_{[0,t)}Z + 1_{[t]}Z_t$

is ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_{t+}}$-measurable. If ${\mathcal{F}_{t+}=\mathcal{F}_t}$ for all t this shows that Z is progressive. Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\lim_{s\downarrow t}X_s = X_t\vee\limsup_{s\downarrow\downarrow t}X_s\smallskip\\ &\displaystyle\lim_{s\rightarrow t}X_s = (\limsup_{s\uparrow t}X_s)\vee(\limsup_{s\downarrow t}X_s) \end{array}$

are also progressive. ⬜

The graph of a random time ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$ is

$\displaystyle [\tau]\equiv\left\{(t,\omega)\in{\mathbb R}_+\times\Omega\colon t=\tau(\omega)\right\}.$

Measurable projection allows us to classify stopping times by the measurability of their graphs.

Lemma 7 Let ${\tau\colon\Omega\rightarrow\bar{\mathbb R}_+}$ be any map. Then,

• ${\tau}$ is measurable if and only if ${[\tau]}$ is jointly measurable.
• ${\tau}$ is a stopping time if and only if ${[\tau]}$ is progressive.

Proof: Clearly, if ${\tau}$ is measurable then ${[\tau]}$ will be jointly measurable and, if ${\tau}$ is a stopping time then

$\displaystyle 1_{[\tau]}=1_{[\tau,\infty)}-1_{(\tau,\infty)}$

is a difference of right and left-continuous adapted processes and, hence, is progressive.

Conversely, setting ${X=1_{[\tau]}}$ then,

$\displaystyle X^*_t=1_{\{\tau\le t\}}.$

So, if X is jointly measurable then Lemma 3 shows that ${\tau}$ is measurable and, if X is progressive, then Lemma 5 shows that ${\tau}$ is a stopping time. ⬜

#### The Debut Theorem

A common use of measurable projection is in the proof of the debut theorem. This states that the first time that a progressively measurable process enters a (Borel) measurable set is a stopping time. We have already proven a version of this for right-continuous processes in these notes, which only required elementary measure theory. I now look at the general version.

The debut of a set ${A\subseteq{\mathbb R}_+\times\Omega}$ is

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle D_A\colon\Omega\rightarrow\bar{\mathbb R}_+,\smallskip\\ &\displaystyle D_A(\omega)=\inf\left\{t\in{\mathbb R}_+\colon(t,\omega)\in A\right\}. \end{array}$

The debut theorem states that this is a stopping time when A is progressive.

Theorem 8 (Debut Theorem) If ${A\subseteq{\mathbb R}_+\times\Omega}$ is progressively measurable, and the underlying filtration is right-continuous, then ${D_A}$ is a stopping time.

Proof: For each positive time t, the set of ${\omega\in\Omega}$ for which ${D_A < t}$ is

$\displaystyle \left\{D_A < t\right\}=\pi_\Omega\left(\left([0,t)\times\Omega\right)\cap A\right)$

As A is progressive ${([0,t)\times\Omega)\cap A}$ is in ${\mathcal{B}({\mathbb R}_+)\otimes\mathcal{F}_t}$ and, by Theorem 1, its projection is in ${\mathcal{F}_t}$. As the filtration is right-continuous, this means that ${D_A}$ is a stopping time. ⬜

Conversely, as ${\{D_A < \infty\}}$ is equal to the projection ${\pi_\Omega(A)}$, the measurable projection theorem is also an immediate consequence of the debut theorem, and Theorems 1 and 8 are seen to be equivalent.

The debut theorem implies that, for progressive processes, hitting times of measurable sets are stopping times. When X is right-continuous and K is closed, this was proven by more elementary methods much earlier in these notes. The generalisation here requires measurable projection.

Theorem 9 If X is a progressively measurable process and ${K\subseteq{\mathbb R}}$ is Borel then, assuming the underlying filtration ${\mathcal{F}_\cdot}$ is right-continuous,

$\displaystyle \tau\equiv\inf\left\{t\in{\mathbb R}_+\colon X_t\in K\right\}$

is a stopping time.

Proof: This follows immediately from Theorem 8, as ${\tau}$ is the debut of the progressively measurable set ${X^{-1}(K)}$. ⬜

#### References

1. Edgar, G. (2014) Most interesting mathematics mistake? MathOverflow. Link.
2. Kanamori, A. (1995) The Emergence of Descriptive Set Theory. From Dedekind to Gödel: Essays on the Deveopment of the Foundations of Mathematics, Sythèse Library volume 251, 241–262. doi:10.1007/978-94-015-8478-4_10. Free pdf available from his website.
3. Lebesgue, H. (1905) Sur les fonctions représentables analytiquement. Journal de Mathématiques Pures et Appliquées. Vol. 1, 139–216. Link.

## 1 Comment »

1. […] aim of this post is to give a fairly direct proof of the theorems of measurable projection and measurable section. These are generally regarded as rather difficult results, and proofs often […]

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