Show $(\mathbb{E}[Z\mid\mathcal{F}_t],\mathcal{F_t})_{t \geq0}$ is an *uniformly integrable* martingale. Suppose $Z \in \mathcal{L}^1(P)$. I want to show that $(\mathbb{E}[Z\mid\mathcal{F}_t],\mathcal{F_t})_{t \geq0}$ is an uniformly integrable martingale.
I have managed to show to it is a martingale but I am stuck trying to show is uniformly integrable as well. I have tried looking at the sufficient condition that
$$
\sup_{t\geq 0} \mathbb{E}[(\mathbb{E}[Z\mid\mathcal{F}_t])^2] < \infty
$$
and my idea was to use Jensens inequality but that doesn't seem to work. I would also need to somehow use that $Z \in \mathcal{L}^1(P)$ as I haven't used that yet.
 A: First, it suffices to consider the case where $Z$ is non-negative. Indeed, write $Z$ as the difference of two integrable non-negative random variables and use the fact that a sum of two uniformly integrable families is uniformly integrable.
For any $t\geqslant 0$ and $R>0$, the event $\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}$ is $\mathcal F_t$ measurable hence
$$
\mathbb E\left[\mathbb E\left[Z\mid\mathcal F_t\right]\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]=\mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right].
$$
Moreover, for each positive $K$,
$$
\mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right]
+K\mathbb P\left(\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}\right)
$$
hence by Markov's inequality,
$$
\mathbb E\left[Z\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right]
+\frac{K}{R}\mathbb E\left[Z\right].
$$
We thus got that for each $R,K>0$,
$$
\sup_{t\geqslant 0}\mathbb E\left[\mathbb E\left[Z\mid\mathcal F_t\right]\mathbf{1}_{\{\mathbb E\left[Z\mid\mathcal F_t\right]>R\}}\right]\leqslant \mathbb E\left[Z\mathbf{1}_{\{Z>K\}}\right]
+\frac{K}{R}\mathbb E\left[Z\right].
$$
Now take $K=\sqrt R$ for example.
A: Uniform integrability may be obtained from the following result.

Proposition: Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space. Suppose that $\frak{A}$ is a collection of $\sigma$--algebras
contained in $\mathscr{F}$.  If $Z\in\mathcal{L}_1(\mathbb{P})$,
then the family $\{\mathbb{E}[Z|\mathscr{A}]: \mathscr{A}\in\frak{A}\}$ is uniformly integrable.

Here is one way to prove this.
Denote $Z_\mathscr{A}:=\mathbb{E}[Z|\mathscr{A}]$. Since
$$|Z_\mathscr{A}|=|\mathbb{E}[Z\,|\mathscr{A}]\,|\leq \mathbb{E}[|Z|\,|\mathscr{A}],$$
$\|Z_\mathscr{A}\|_1=\mathbb{E}[|Z_\mathscr{A}|]\leq \|Z\|_1=\mathbb{E}[|Z|]$, and
\begin{align}
\int_{\{|Z_\mathscr{A}|>a\}}|Z_\mathscr{A}|\,d\mathbb{P}\leq 
\int_{\{\mathbb{E}[|Z|\,|\mathscr{A}]>a\}}\mathbb{E}[|Z|\,|\mathscr{A}]\,d\mathbb{P}=
\int_{\{\mathbb{E}[|Z|\,|\mathscr{A}]>a\}} |Z|\,d\mathbb{P}.
\end{align}
Since $\mathbb{P}\big(\mathbb{E}[|Z|\,|\mathscr{A}]>a\big)\leq\frac{\mathbb{E}[|Z|]}{a}\xrightarrow{a\rightarrow\infty}0$,
we conclude that
$$\inf_{a>0}\sup_{\mathscr{A}\in\frak{A}}
\int_{\{|Z_{\mathscr{A}}|>a\}}|Z_\mathscr{A}|\,d\mathbb{P} =0.$$
Hence $\{Z_\mathscr{A}:\mathscr{A}\in\frak{A}\}$ is uniformly integrable.
