Proof for a maximal inequality without using optional stopping theorems Is it possible to prove the following maximal inequality without using optional stopping theorems or any result related to stopping times:
For a nonnegative sub martingale $X_n$ $(\infty$ is a possible value$)$$,\lambda>0$ $$\lambda P(\max_{1 \leq k \leq n}X_k>\lambda)\leq E[X_n1_{\{\max_{1 \leq k \leq n} X_k>\lambda\}}] \ \ \ \ \ \ \ \ \ (1)$$
Considering $F(x)=(x-\lambda)^+$ and $f(x)=1_{[\lambda,+\infty[}(x),$ so $F$ is convex, increasing and $F'=f$ (we can consider the derivative at right).
Let $Y_n=\max_{1 \leq k \leq n}X_k,W_n=F(Y_n)-(Y_n-X_n)f(Y_n).$ So $Y_n$ is a sub martingale, since $F(Y_{n+1})-F(Y_n)-(Y_{n+1}-X_{n+1})f(Y_{n+1})+(Y_n-X_n)f(Y_n)=F(Y_{n+1})-F(Y_n)-(Y_{n+1}-X_{n+1})f(Y_{n})+(Y_n-X_n)f(Y_n) \geq f(Y_{n})(Y_{n+1}-Y_n)-f(Y_n)(Y_{n+1}-X_{n+1})+(Y_n-X_n)f(Y_n) \geq f(Y_n)(X_{n+1}-X_n).$
This means that $0 \leq E[F(X_1)] \leq E[F(Y_n)]-E[f(Y_n)(Y_n-X_n)],$ so $$\lambda P(Y_n>\lambda) \leq E[X_n1_{\{Y_n>\lambda\}}].$$ In order to have finite values (integrals are well defined...), tried to truncate: $V_n=\min(k,X_n),$ but $V_n$ is not necessarily a sub martingale.
Do you any other ways to prove $(1)$? Also in the above way, how to deal with the integrability (especially infinite values)?
 A: Supermartingales are generally assumed to have finite first moments in which case your proof is complete. However, since you specified that these supermartingales can reach infinity, we do need to consider the non-integrable state separately. This is actually quite simple. First, note that whenever $Y_n = \infty$, $X_n = \infty$. Thus, if there exists an $n$ such that $P(X_n = \infty) > 0$, then
$$E[X_n\mathbb{I}_{Y_n > \lambda}] = \infty.$$
In this case, the inequality holds trivially. So, we can assume without loss of generality that $X_n$ is almost surely finite. In this case, your reasoning holds even when $E[|X_n|] = \infty$, so $W_n$ is a submartingale (minus the finite first moment condition). Now note that we can alternatively characterize $W_n$ by,
$$W_n = \begin{cases}
X_n - \lambda &\text{ if } Y_n > \lambda,\\
0 &\text{ if } Y_n \leq \lambda.
\end{cases} = (X_n - \lambda)\mathbb{I}_{Y_n > \lambda}$$
It follows that $W_n$ is a submartingale and $E[W_1] = E[(X_1 - \lambda)^+] \geq 0$. Then,
$$E[W_n] = E[X_n\mathbb{I}_{Y_n > \lambda}] - \lambda P(Y_n > \lambda) \geq E[W_1] = 0.$$
Now, if $E[W_n] < \infty$, then your arguments hold and we get the inequality. However, note that $\lambda P(Y_n > \lambda) \leq \lambda < \infty$. So, if $E[W_n] = \infty$, then
$$E[X_n\mathbb{I}_{Y_n > \lambda}] - \lambda P(Y_n > \lambda) = \infty \Rightarrow E[X_n\mathbb{I}_{Y_n > \lambda}] = \infty.$$
So $\lambda P(Y_n > \lambda) < E[X_n\mathbb{I}_{Y_n > \lambda}] = \infty.$
