The usual optimal stopping theorem states the following: Let $(\Omega, \mathcal{B}, \mathbb{P})$ be a probability space. Let $\{X_n\}$ be a super martingale and $T$ be a stopping time with respect to a filtration $\{\mathcal{B}_n\}$ in $\mathcal{B}.$ Then $E[X_T] \leq E[X_1]$ whenever the following conditions hold:
1) $\Pr\{T < \infty\} = 1.$
2) $\mathbb{E}(|X_T|) < \infty.$
3) $\lim_{m\rightarrow\infty}\mathbb{E}(|X_m|1_{T > m}) = 0.$
However the following proof 'seems' to indicate that only condition 1 is necessary. I know this proof should be incorrect (as standard examples show), but I am unable to identify the mistake. So kindly help me with the same.
Let $Y_n = X_{n\wedge T}$ $\forall n.$ Clearly, $\{Y_n\}$ is a super martingale. Since $\Pr\{T < \infty\} = 1,$ it also follows that $Y_n \rightarrow X_T$ a.s. Now by monotonicity property, it follows that
\begin{equation} \mathbb{E}\left(\bigwedge_{m \geq k}Y_m|\mathcal{B}_1 \right) \leq \mathbb{E}(Y_k|\mathcal{B}_1) \hspace{0.2cm} \forall k \end{equation}
But because $\{Y_n\}$ is a super martingale, $\mathbb{E}(Y_k|\mathcal{B}_1) \leq Y_1 = X_1\wedge T = X_1$ (since $T\geq 1).$ It thus follows that
\begin{equation} \mathbb{E}\left(\bigwedge_{m \geq k}Y_m|\mathcal{B}_1 \right) \leq X_1. \end{equation}
Now by monotone convergence theorem for conditional expectations,
\begin{equation} \mathbb{E}(X_T|\mathcal{B}_1) = E\left(\lim_{k \rightarrow \infty}\bigwedge_{m \geq k}Y_m|\mathcal{B}_1\right) = \lim_{k \rightarrow \infty} \mathbb{E} \left(\bigwedge_{m \geq k}Y_m |\mathcal{B}_1\right) \leq X_1. \end{equation}
Taking expectations again, we get that
\begin{equation} \mathbb{E}(X_T) \leq \mathbb{E}(X_1) \end{equation} as desired.