Uniform Integrability of $M^2$ if $M$ is a martingale. Let $M:\Omega\times\mathbb{R^+}\longrightarrow \mathbb{R}$ be a martingale such that
\begin{equation*}
\sup_{t\in \mathbb{R}^+}\mathbb{E}[M_t^2]<\infty
\end{equation*}
Then is $M$ uniformly integrable? i.e. is it true that
\begin{equation*}
\lim_{K\rightarrow\infty} \left[ \sup_{t\in \mathbb{R}^+} \mathbb{E}\left[ M_t^2 \mathbb{I}_{\left\{M_t^2> K\right\}}\right]\right] =0
\end{equation*}
I think it is...and my attempt at proving it so far is as follows:

However I am getting stuck at the last step where I need to interchange the limit and the supremum...(I have highlighted this step in red). I know that we can interchange the limit and the suprermum if $\sup_{t\in \mathbb{R}^+} \mathbb{E}\left[ M_t^2 \mathbb{I}_{\left\{M_t^2> K\right\}}\right]$ converges uniformly to 0 when we take the limit as $C\rightarrow\infty$, but I couldn't show this either....
Any help would be greatly appreciated, whether it be:
1) showing another way of proving it, 
2) showing a way of getting around/proving the last step
3) telling me this statement isn't true (and if it isn't is there a counter example).
Many thanks!
 A: As pointed out by carmichael561, a family of random variable with is bounded in $\mathbb L^p$ is uniformly integrable. Note that the post addresses the case of sequences, but the same proof works for a family of random variables with an arbitrary index. This would be sufficient to prove uniform integrability of $\{\left\lvert M_t\right\rvert,t\geqslant 0\}$ but not that of $\{  M_t^2,t\geqslant 0\}$. 
Here are the ideas to handle it. 


*

*It suffices to prove the uniform integrability of $\{M_n^2,n\in\mathbb N\}$. 
Let $n$ be a positive integer and let $t\in [n-1,n)$. Then for all positive $x$, 
\begin{align}
\Pr\left(\left\lvert M_t\right\rvert>x\right)&=\Pr\left(\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert>x\right)\\
&\leqslant \frac 1x\mathbb E\left[\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert\mathbf 1\left\{\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert>x\right\}   \right]\\
&\leqslant \frac 1x\mathbb E\left[
\mathbb E\left[\left\lvert 
M_n\right\rvert\mid\mathcal F_t\right]\mathbf 1\left\{\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert>x\right\}   \right].
\end{align}
Since the set $\left\{\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert>x\right\}$ belongs to $\mathcal F_t$, we obtain by definition of the conditional expectation that 
$$\Pr\left(\left\lvert M_t\right\rvert>x\right)\leqslant \frac 1x\mathbb E\left[
 \left\lvert 
M_n \right\rvert\mathbf 1\left\{\left\lvert 
\mathbb E\left[
M_n\mid\mathcal F_t\right]\right\rvert>x\right\}   \right].$$
Observing that for all non-negative integrable random variable $X$ and any event $A$, 
$$
\mathbb E\left[X\mathbf 1_A\right]=\int_0^{+\infty}\Pr\left(\{X>t\}\cap A\right)dt
\leqslant \frac x2\Pr(A)+\int_{x/2}^{+\infty}\Pr\left( X>t\right)dt\leqslant 
 \frac x2\Pr(A)+\frac 2x\mathbb E\left[\mathbf 1\{X>x/2\}\right],
$$
we get 
$$\Pr\left(\left\lvert M_t\right\rvert>x\right)\leqslant \frac 4x\mathbb E\left[
 \left\lvert 
M_n \right\rvert\mathbf 1\left\{\left\lvert 
M_n\ \right\rvert>x/2\right\}   \right].$$
To conclude, express $\mathbb E\left[M_t^2\mathbf 1\left\{\left\lvert 
M_t  \right\rvert>R\right\}\right]$ in term of the tail of $\left\lvert 
M_t  \right\rvert$.


*In order to show the uniform integrability of $\{M_n^2,n\in\mathbb N\}$, we apply Doob's inequality in order to see that $\sup_{n\in\mathbb N}M_n^2$ is integrable.



The concerns about the approach proposed in the opening post are the following:


*

*In the image, after the sentence "we may conclude that", the first equality is obvious ($C$ and $K$ play the same role).

*The red equality needs justifications, since it is in general false that we can exchange a limit and a supremum. 

