I am trying to prove the Doob's Upcrossing Lemma and the first step requires to prove that: If $X$ is a submartingale, then $(X-a)_+$ is a submartingale. I found it intuitive but i failed to prove. Here is my attempt:
$\mathbb{E}[(X_{n+1}-a)_+|\mathscr{F_n}]\\=\mathbb{E}[(X_{n+1}-a)|\mathscr{F_n}]+\mathbb{E}[(X_{n+1}-a)_-|\mathscr{F_n}]\\>(X_{n}-a)+\mathbb{E}[(X_{n+1}-a)_-|\mathscr{F_n}]\\=(X_{n}-a)_+-(X_{n}-a)_-+\mathbb{E}[(X_{n+1}-a)_-|\mathscr{F_n}]\\=(X_{n}-a)_++(\mathbb{E}[(X_{n+1}-a)_-|\mathscr{F_n}]-(X_{n}-a)_-)$
If $(X_{n}-a)_-$ is a martingale or submartingale this is fine. What if supermartingale?
Notice that $(X_{n+1 }-a) ^+\geqslant X_{n+1}-a $, hence taing the conditional expectation with respect to $\mathcal F_n$ yields $$\mathbb E\left[(X_{n+1 }-a) ^+\mid\mathcal F_n\right] \geqslant \mathbb E\left[(X_{n+1 }-a)\mid\mathcal F_n\right].$$ Now find a below bound using the submartingale property.