# Is it true that $E[|Y|\mid{\mathcal G}]\leq |Y|$ almost surely?

This is an exercise of conditional expectations:

Let $Y$ be an integrable random variable on the space $(\Omega,{\mathcal A},{\bf P})$ and $\mathcal{G}$ be a sub $\sigma$-algebra of $\mathcal{A}$. Show that $|Y|\leq c$ implies $|E[Y\mid{\mathcal G}]|\leq c$.

With Jensen's inequality, one immediately has $|E[Y\mid{\mathcal G}]|\leq E(|Y|\mid{\mathcal G})$.

I am trying to show that $E[|Y|\mid{\mathcal G}]\leq |Y|$, which is not necessarily true though. If $Y$ is $\mathcal{G}$-measurable, then $E[|Y|\mid {\mathcal G}]= |Y|$. But I have no idea for the general case.

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Hint: Choose $\mathcal G = \{\varnothing, \Omega\}$. – cardinal Nov 29 '12 at 1:25
@cardinal I don't see your point. – Jack Nov 29 '12 at 2:14
Ok. With the given $\mathcal G$, what is $\mathbb E( |Y| \mid \mathcal G)$ equal to? Now, think about what that means and consider a simple case, say, e.g., a discrete nonnegative $Y$. – cardinal Nov 29 '12 at 2:30

The only case when $\mathbb E(Z\mid\mathcal G)\leqslant Z$ almost surely, for some integrable random variable $Z$ and some sigma-algebra $\mathcal G$, is when $Z$ is measurable with respect to $\mathcal G$. Then $\mathbb E(Z\mid\mathcal G)=Z$ almost surely.
Proof: Call $T=\mathbb E(Z\mid\mathcal G)$, then $\mathbb E(T)=\mathbb E(Z)$. Hence, if $T\leqslant Z$ almost surely, then $T=Z$ almost surely. If this happens, $Z$ is measurable with respect to $\mathcal G$ because $T$ is, by definition of conditional expectation.