## Lecture 2. Measure theory in function spaces

Stochastic processes can be seen as random variables taking their values in a function space. It is therefore important to understand the naturallly associated $\sigma$-algebras.

Let $\mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d)$, $d \ge 1$, be the set of functions $\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}^d$. We denote by $\mathcal{T}(\mathbb{R}_{\ge 0},\mathbb{R}^d)$ the $\sigma$-algebra generated by the so-called cylindrical sets

$\{ f \in \mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d), f(t_1) \in I_1,...,f(t_n) \in I_n \}$

where $t_1,...,t_n \in \mathbb{R}_{\ge 0}$ and where $I_1,...,I_n$ are products of intervals: $I_i=\Pi_{k=1}^d (a^k_i,b^k_i]$.
As a $\sigma$-algebra $\mathcal{T}(\mathbb{R}_{\ge 0},\mathbb{R}^d)$ is also generated by the following families:

• $\{ f \in \mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d ), f(t_1) \in B_1,...,f(t_n) \in B_n \}$where $t_1,...,t_n \in \mathbb{R}_{\ge 0}$ and where $B_1,...,B_n$ are Borel sets in $\mathbb{R}^d$.
• $\{ f \in \mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d), (f(t_1),...,f(t_n)) \in B \}$where $t_1,...,t_n \in \mathbb{R}_{\ge 0}$ and where $B$ is a Borel set in $(\mathbb{R}^{d})^{\otimes n}$.

Exercise: Show that the following sets are not in $\mathcal{T} ([0,1],\mathbb{R})$:

• $\{ f \in \mathcal{A}([0,1], \mathbb{R}), \sup_{t\in [0,1]} f(t) <1 \}$
• $\{ f \in \mathcal{A}([0,1], \mathbb{R}), \exists t\in [0,1] f(t) =0 \}$

The above exercise shows that the $\sigma$-algebra $\mathcal{T}(\mathbb{R}_{\ge 0},\mathbb{R}^d)$ is not rich enough to include natural events; this is due to the fact that the space $\mathcal{A}(\mathbb{R}_{\ge 0}, \mathbb{R}^d)$ is by far too big.

In these lectures, we shall mainly be interested in processes with continuous paths. In that case, we use the space of continuous functions $\mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R}^d)$ endowed with the $\sigma$-algebra $\mathcal{B}(\mathbb{R}_{\ge 0},\mathbb{R}^d)$ that is generated by

$\{ f \in \mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R}^d), f(t_1) \in I_1,...,f(t_n) \in I_n \}$

where

$t_1,...,t_n \in \mathbb{R}_{\ge 0}$

and where $I_1,...,I_n$ are products of intervals $\Pi_{k=1}^d (a^k_i,b^k_i]$. This $\sigma$-algebra enjoys nice properties. It is for instance generated by the open sets of the (metric) topology of uniform convergence on compact sets.

Proposition
The $\sigma$-algebra $\mathcal{B} ( \mathbb{R}_{\ge 0}, \mathbb{R}^d)$ is generated by the open sets of the topology of uniform convergence on compact sets.

Proof:

We make the proof when the dimension $d=1$ and let the reader adapt it in higher dimension. Let us first recall that, on $\mathcal{C}(\mathbb{R}_{\ge 0},\mathbb{R})$ the topology of uniform convergence on compact sets is given by the distance

$d(f,g)=\sum_{n=1}^{+\infty} \frac{1}{2^n} \min (\sup_{0 \le t \le n} \mid f(t) -g(t) \mid ,1).$

This distance endows $\mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R})$ with the structure of a complete, separable, metric space (that is of a Polish space). Let us denote by $\mathcal{O}$ the $\sigma$-field generated by the open sets of this metric space.

First, it is clear that the cylinders

$\{ f \in \mathcal{C} (\mathbb{R}_{\ge 0}, \mathbb{R}), f(t_1) < a_1,...,f(t_n) < a_n \}$

are open sets that generate $\mathcal{B} (\mathbb{R}_{\ge 0},\mathbb{R})$. Thus, we have

$\mathcal{B} (\mathbb{R}_{\ge 0}, \mathbb{R}) \subset \mathcal{O}.$

On the other hand, since for every $g \in \mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R})$, $n \in \mathbb{N}$, $n \geq 1$ and $\rho >0$,

$\{ f \in \mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R}), \sup_{0 \le t \le n} \mid f(t) -g(t) \mid \le \rho \}$

$=\cap_{t \in \mathbb{Q}, 0 \le t\le n} \{ f \in \mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R}), \mid f(t)-g(t) \mid \le \rho \},$

we deduce that

$\{ f \in \mathcal{C}(\mathbb{R}_{\ge 0}, \mathbb{R}), \sup_{0 \le t \le n} \mid f(t) -g(t) \mid \le \rho \} \in \mathcal{B} (\mathbb{R}_{\ge 0}, \mathbb{R}).$

Since $\mathcal{O}$ is generated by the above sets, this implies

$\mathcal{O} \subset \mathcal{B} (\mathbb{R}_{\ge 0}, \mathbb{R})$

and concludes the proof. $\square$

Exercise.
Show that the following sets are in $\mathcal{B} ([0,1],\mathbb{R})$:

• $\{ f \in \mathcal{C}([0,1], \mathbb{R}), \sup_{t\in [0,1]} f(t) <1 \}$
• $\{ f \in \mathcal{C}([0,1], \mathbb{R}), \exists t\in [0,1] f(t) =0 \}$
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### 5 Responses to Lecture 2. Measure theory in function spaces

1. zihui says:

Thank you Professor Baudoin for your wonderful website. May I ask a question. When defining the cylindrical sets, why do you have to use multiple n? If I define the cylindrical sets the same way as yours but only for n=1, wouldn’t they still generate the same sigma algebra? Thanks.

• That’s a very good remark. If we consider cylindrical sets with n=1, they would actually generate the very same sigma-algebra because the cylinder with n >1 is an intersection of a finite number of cylinders with n=1.

• zihui says:

Professor Baudoin, thank you very much for your reply. I am thinking about the first two exercise problems. Generally it is easier to show something is an element of a sigma-algebra, but how can I argue something is not in a sigma-algebra? I have a feeling that an element of the sigma-algebra T only has something to say for values of the functions on countably many t’s in [0,1], while sup f(t) < 1 poses restrictions on uncountably many t's. But how do I make the argument rigorous? Could you give a short hint?

2. zihui says:

We can define the distance d the same way on both spaces C and A (but then C is a Polish spaces while A is not, because A is not separable). Denote the sigma-algebras generated by cylinder sets in the two spaces by B and T respectively, then B is precisely the Borel sigma-algebra in C but T is not the Borel sigma-algebra in A (therefore we say B has nice properties in C while T is too small in A). Am I correct?

3. Nikita Evseev says:

There are two typos in words “naturallly” and “indude”. And in the proof you use the term sigma-field instead of sigma-algebra.