Does a property derived using optional stopping hold in the limit

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Let $(X_n)_n$ be a random walk with a stopping time $\tau_c=\min\{n\geq 0:X_n\leq 0 \text{ or } X_n\geq c\}$ where $c>0$ is a parameter. Assume that I manage to show that $\mathbb{E}(\tau_c)\geq K$ $\forall c>\mu$ where $K$ is independent of $c$ and $\mu$ is some cutoff.

Can I then conclude that the stopping time $\tau=\min\{n\geq 0:X_n\leq 0\}$ also fulfills $\tau\geq K$?

I think that this is true since I'd say that $\tau=\lim_{c\rightarrow\infty}\tau_c$. But I like to be careful when I take my limits.

I am especially careful here since I am actually only interested in $\tau$, but without the upper bound $c$ I cannot prove the inequality. With the bound (which has to be finite but can otherwise be arbitrarily large) I can use Doob's optional stopping theorem and prove that the lower bound on the stopping times holds. This is why I am hoping that I can introduce the bound $c$, show that my lower bound $K$ holds for all $c$ and then conclude that it must also hold for $c\rightarrow\infty$.