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Let $Z$ be an integrable random variable on filtered probability space $(\Omega , \mathscr F, (\mathscr {F_n})_{\{n \in \mathbb{N}\}}, \mathbb P)$
Define $Z_{n} := E[Z|\mathscr {F_n}]$. Show that $Z_n$ is integrable i.e. $E[|Z_n|] < \infty$
What I tried:
$E[|Z_n|] = E[|E[Z|\mathscr {F_n}]|] \le E[E[|Z||\mathscr {F_n}]] = E[|Z|] < \infty$ QED
Did I do that inequality* right? Any mistakes?
*It's the Triangle Inequality for Integrals.
You could invoke Jensen's inequality for conditional expectation. Here's an easier way to prove it, in the sense that it requires less background machinery.
For a random variable $X$ let $X^- = \max(-X, 0)$ and $X^+ = \max(X, 0)$ be its negative and positive parts. By definition $\mathbb E X = \mathbb E X^+ - \mathbb EX^{-}$ whenever at least one of the terms is finite. Thus, $\mathbb E X < \infty$ if and only if $ \mathbb E X^+<\infty$ and $ \mathbb E X^- < \infty$. Thus, $\mathbb E X <\infty \iff \mathbb E X^+ + \mathbb EX^{-} = \mathbb E \vert X \vert <\infty$. But $\mathbb E Z_n = \mathbb E Z <\infty$ so indeed $\mathbb E \vert Z_n \vert <\infty$, which is what we wanted to prove.
The math you stated looks right. I'm not aware of any name for the inequality, but just from the title it is fairly intuitive. If you had any (finite) set of points, any non-finite value would cause the expectation to be non-finite.
