Right-Limits of Filtration of Sigma-Algebra form Filtration

Theorem

Let $\struct {X, \Sigma}$ be a measurable space.

Let $\sequence {\FF_t}_{t \ge 0}$ be a continuous-time filtration of $\Sigma$.

For each $t \in \hointr 0 \infty$, let $\FF_{t^+}$ be the right-limit of $\sequence {\FF_t}_{t \ge 0}$ at $t$.

Then $\sequence {\FF_{t^+} }_{t \ge 0}$ is a filtration of $\Sigma$.

Proof

We need to show that $\FF_{t^+}$ is a sub-$\sigma$-algebra of $\Sigma$ for each $t \in \hointr 0 \infty$, and:

$\FF_{t^+} \subseteq \FF_{q^+}$

for $t, q \in \hointr 0 \infty$ with $t \le q$.

From the definition of the right-limit, we have:

$\ds \FF_{t^+} = \bigcap_{s > t} \FF_s$

for each $t \in \hointr 0 \infty$.

From Intersection of Sigma-Algebras, we have that $\FF_{t^+}$ is a sub-$\sigma$-algebra of $\Sigma$ for each $t \in \hointr 0 \infty$.

Let $t, q \in \hointr 0 \infty$ with $t \le q$.

Then, we have:

$\set {s \in \R : s > q} \subseteq \set {s \in \R : s > t}$

and so:

$\ds \bigcap_{s > t} \FF_s \subseteq \bigcap_{s > q} \FF_s$

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so that:

$\FF_{t^+} \subseteq \FF_{q^+}$

$\blacksquare$