Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $\tau:L^1(\operatorname P)\to L^1(\operatorname P)$ be linear and continuous with respect to the weak topology on $L^1(\operatorname P)$. Let $Z\in L^1(\operatorname P)$ be a limit point of $(Z_n)_{n\in\mathbb N}\subseteq L^1(\operatorname P)$ with respect to the weak topology on $L^1(\operatorname P)$ and assume that $$\tau Z_n\le\frac Xn+Y\tag2$$ for some $X,Y\in L^1(\operatorname P)$.
Why can we conclude that $\tau Z\le Y$?
The continuity assumption should be equivalent to $$\operatorname E[\tau(U_n)V]\xrightarrow{n\to\infty}\operatorname E[\tau(U)V]\;\;\;\text{for all }V\in L^\infty(\operatorname P)\cong L^1(\operatorname P)'\tag2$$ for all $(U_n)_{n\in\mathbb N}\subseteq L^1(\operatorname P)$ and $U\in L^1(\operatorname P)$ with $\operatorname E[U_nV]\xrightarrow{n\to\infty}\operatorname E[UV]$ for all $V\in L^\infty(\operatorname P)$, if I'm getting things right.
Suppose that we do not have that $\tau Z \leq Y$ a.e. Then there is an $\varepsilon > 0$ and a set $A$ of positive measure such that $\tau Z \geq Y + \varepsilon$ on $A$ (by writing $\{\tau Z > Y\}$ as the countable union of the sets $\{ \tau Z \geq Y + n^{-1}\}$).
We can pass to a subnet $(Z_\lambda)$ of the sequence $Z_n$ to assume that $Z_\lambda \to Z$ in the weak sense as a net. This means that for any $V \in L^\infty$, we have $E[ Z_\lambda V] \to E[Z V]$. By taking $V = 1_A$, we can then write $$\int_A Y d\mathbb{P} + \varepsilon \mathbb{P}(A) \leq \int_A \tau Z d\mathbb{P} = \lim_{\lambda} \int_A \tau Z_\lambda d\mathbb{P} \leq \lim_{\lambda} \int_A \frac{X}{\lambda} + Y d\mathbb{P} = \int_A Y d \mathbb{P}$$ which is a contradiction.