I am not sure that this is a proper question. I am looking for the verification of the proof rather than for an answer.
There is one inequality I use pretty often and I would like to be sure that it is correct. Namely, if $X_n$ is a non-negative supermartingale then $ \mathsf P\left\{\sup\limits_n X_n\geq \delta\right\}\leq\frac 1 \delta\mathsf EX_0.\quad(1) $
First I derived it from the version of Doob's inequality presented in the book "Optimal Stopping and Free-Boundary Problems" by Peskir and Shiryaev (2006). On the other hand here Doob's inequality in probability you can see a counterexample for the inequality stated in the book. Also it that topic mpiktas find a reference to the earlier book by Shiryaev with a weaker inequality, which I believe is correct.
Anyway, I have a proof of (1) and would like you to help me verify if it is strict or not.
Let $Y$ be a process with non-negative values. Then for all $N\in \mathbb{N}$ and $\delta>0$ holds $ \mathsf P\left\{\sup\limits_{n\leq N} Y_n\geq \delta\right\}\leq\frac 1 \delta\sup\limits_{\tau\leq N}\,\,\,\mathsf EY_\tau\quad (2) $ where $\tau$ is a stopping time with respect to the natural filtration of $Y$.
Proof: Take $\tau^*=\inf\{ n\geq 0:Y_n\geq\delta\}$ and $\tau = \tau^*\wedge N$. Now $\tau\leq N$ a.s. and $\tau$ is a stopping time with respect to the natural filtration of $Y$. Denote $p = \mathsf P\left\{\sup\limits_{n\leq N} Y_n\geq \delta\right\},$ so now $\mathsf{P}\{\tau = \tau^*\} = p$.
We have $ \mathsf EY_\tau \geq \delta p $ since $Y$ is a non-negative process, which leads to the inequality (2). End of the proof.
We can obtain (1) from (2) since $\sup\limits_{\tau\leq N} \,\,\,\mathsf EX_\tau = \mathsf E X_0$ if $X$ is a supermartingale. Taking $N\to\infty$ we obtain (1).