# Measurable funtion and sub sigma algebra

If $X_t$ is measurable w.r.t a sigma algebra $F_t$, what is the relationship between $X_t$ and a sub-sigma-algebra $F_{t_1}$ s.t. $F_{t_1}$ is included in $F_t$ ?

A preliminary thought question, Thanks.

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I would say that there is no useful relationship.

Let $(X, \Omega)$ and $(Y, \Psi)$ be measure space where $\Omega$ and $\Psi$ are $\sigma$-algebras on $X$ and $Y$, respectively. Let $f : (X, \Omega) \rightarrow (Y, \Psi)$ be a measurable function. This means that $f^{-1}(B) \in \Omega$ for all $B \in \Psi$.

Now suppose that $X = Y$ and $X$ is a infinite set. Let $\Omega = \Psi = \mathcal{P}(X)$, the power set. Let $\Omega_1 = \{0, X\}$ so that $\Omega_1 \subset \Omega$. Let $f : X \rightarrow X$ be the identity function. Then $f : (X, \Omega) \rightarrow (X, \Omega)$ is a measurable function. However, $f : (X, \Omega_1) \rightarrow (X, \Omega)$ is not.

On the other hand, if $X = Y$ and $\Omega = \mathcal{P}(X)$, $\Psi = \{0, X\}$ and $\Omega_1 = \{0, X\}$. Letting $f$ be the identity function, $f : (X, \Omega) \rightarrow (X, \Psi)$ is a measurable. And $f : (X, \Omega_1) \rightarrow (X, \Psi)$ is still measurable.

So, it possible that a measurable function remains measurable or becomes not measurable when restricted to sub-$\sigma$-algebras.

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The conditional expectation of X with respect to the sub-σ-field is a measurable function of X.

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No, not in general. –  Did May 10 '14 at 10:29