I am trying to prove the $L^p$ Martingale convergence theorem for martingale $X=(X_n)^{\infty}_{n=0}$ on $(\Omega,\mathcal{F},(\mathcal{F}_n)^\infty_{n=0},\mathbb{P})$ which is bounded in $L^p$ for some $p>1$. I followed the example in my textbook to divide the cases for $X_\infty$ is bounded and unbounded. I have problem in understanding the unbounded case which is as follow: $$\mathbb{E}[|X_n-X_\infty|^p]^{1/p} = \mathbb{E}[|\mathbb{E}[X_\infty|\mathcal{F_n}]-X_\infty|^p]^{1/p}\\ \leq\mathbb{E}[|\mathbb{E}[X_\infty1_{|X_\infty|\leq K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|\leq K}|^p]^{1/p}+\mathbb{E}[|\mathbb{E}[X_\infty1_{|X_\infty|> K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|> K}|^p]^{1/p}\\\leq\mathbb{E}[|\mathbb{E}[X_\infty1_{|X_\infty|\leq K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|\leq K}|^p]^{1/p}+2\mathbb{E}[|X_\infty1_{|X_\infty|> K}|^p]^{1/p}$$
In the second inequality, why $$\mathbb{E}[|\mathbb{E}[X_\infty1_{|X_\infty|> K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|> K}|^p]^{1/p}<2\mathbb{E}[|X_\infty1_{|X_\infty|> K}|^p]^{1/p}$$ and is $X_n1_{|X_n|>=K}$ a martingale?
The inequality follows from the fact that if $a$ and $b$ are two real numbers, then (by convexity of $x\mapsto |x|^p$, $$|a-b|^p\leqslant (|a|+|b|)^p\leqslant 2^{p-1}(|a|^p+|b|^p),$$ hence by conditional Jensen's inequality, $$|\mathbb{E}[X_\infty1_{|X_\infty|> K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|> K}|^p \leqslant 2^{p-1}\left(\mathbb E[|X_\infty|^p\mathbb 1_{|X_\infty\gt K}\mid\mathcal F_n]+|X_\infty|^p\mathbb 1_{|X_\infty\gt K}\right).$$ Taking the expectation yields $$\mathbb E|\mathbb{E}[X_\infty1_{|X_\infty|> K}|\mathcal{F_n}]-X_\infty1_{|X_\infty|> K}|^p\leqslant 2\mathbb E\left[|X_\infty|^p\mathbb 1_{|X_\infty\gt K}\right].$$
To conclude, we use the result for bounded sequence with the bounded martingale $\left(\mathbb E[X_\infty\mathbb 1_{|X_\infty\leqslant K}\mid\mathcal F_n]\right)$.