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.

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