For a martingale $M$, $\lim_{a\to\infty}\lim_{t\to\infty} \mathbf E( M_t; \int_0^t M_s ds > a ) = 0$?

Question

Does the following equality right? \begin{equation} \lim_{a\to\infty}\lim_{t\to\infty} \mathbf E\left( M_t; \int_0^t M_s ds > a \right) = 0, \end{equation} where $M = \{M_t\}_{t\ge0}$ is a martingale.

Intuitively, I think it is right. Because $\mathbf E(M_t) = \mathbf E(M_0)$ is a finite time-independent constant. And as $a\to\infty$, $\mathbf P(\int_0^t M_s ds > a)$ is expect to vanish.

But I don't know how to prove it rigorously. Can anyone give some hints? Or if the equality does not hold true, some counterexamples will be appreciated. TIA...

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PS: By $\mathbf E(X;A)$, I mean $\mathbf E(X\mathbf 1_A)$.

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EDIT: As answered by @user159517 through a trivial counterexample, this is wrong in general. However, is there any possibility that it does hold for some non-trivial martingales, say, for example, the exponential martingale which is crucial in the Girsanov theorem?

Answer 1

This is wrong. Let $M_t \equiv 1$, then $$\mathbb E\left( M_t; \int_0^t M_s ds > a \right) = \mathscr{1}_{[\int_0^t M_s ds > a]} = \mathscr{1}_{[t>a]}$$ It follows that $$ \lim_{a\to\infty}\lim_{t\to\infty} \mathbf E\left( M_t; \int_0^t M_s ds > a \right) = 1.$$