Is it true $P(\sup_{k \in \mathbb{N}}X_k \geq \epsilon +x)=\dfrac{x}{\epsilon+x}$? Let $(X_k)_k$ be a non-negative martingale such that $\lim_{k \to \infty}X_k=0$ a.s. and $X_0=x\in ]0,\infty[.$
Prove or disprove the following: $$\forall\epsilon>0,P(\sup_{k \in \mathbb{N}}X_k \geq \epsilon +x)=\frac{x}{\epsilon+x} \ \ \ \ \ \ \ \ \ (1)$$
I deduced, from a maximal inequality, that $P(\sup_{k \in \mathbb{N}}X_k \geq \epsilon +x) \leq \dfrac{E[X_0]}{x+\epsilon}=\dfrac{x}{\epsilon+x}.$
Any suggestions how to prove $(1)?$

In general, (1) may not hold: take $X_k=x\prod_{i=1}^k\left(\mathbf{1}_{A_i}/p_i\right)$ where $(A_i)$ is an independent sequence of events, $p_i=\mathbb P(A_i)>0$ and $\mathbb P\left(\bigcap_{i\geqslant 1}A_i\right)=0$.
Then the sequence $(X_k)$ can take only the values $x$, $0$, $1/p_1$, $1/(p_1p_2)$, $\prod_{i=1}^k 1/p_i$, $k\geqslant 1$ and so does $\sup_{k\in\mathbb N}X_k$, as $\prod_{i=1}^k 1/p_i\to \infty$.
