$\mathbb{E}[M_{a}]\geq \mathbb{E}[M_{b}]$ with $a,b$ limited stopping times and $M$ sub-martingale

Question

I don't understand why $$\mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}X_T\mid \Im_t]\geq \mathbb{E}^{\mathbb{Q}}[e^{-r(\tau-t)}X_{\tau}\mid \Im_t] \Rightarrow \mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}X_T\mid \Im_t]\geq \sup_{\tau}\mathbb{E}^{\mathbb{Q}}[e^{-r(\tau-t)}X_{\tau}\mid \Im_t]$$

Let $\begin{Bmatrix} X_t \end{Bmatrix}_{t\in [0,+\infty)}$ be a sub-martingale $\Im_t$-adapted, with $\tau:=(t+s) \leq T \space \mathbb{P}$-$ \operatorname{a.s.}$ for $T$ generical stopping time.

Now, if $$\sup_{\tau}\mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}X_T\mid \Im_t]=\mathbb{E}^{\mathbb{Q}}[e^{-r(T-t)}X_T\mid \Im_t]$$ because the variable $\tau$ is not the variable of interest...

...why if $\tau \leq T$ also $\sup_{\tau} \leq T$?

Thanks in advance.