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Let's say I have a random function $X(t)$ that is continuous in $t$, almost surely.

Is it true that $$\mathbb E(X(t_1)) = \mathbb E\left(\lim_{t\to t_1} X(t)\right)?$$

This seems incorrect to me but I have no clue... Thanks!

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  • $\begingroup$ In general, no. See the limit theorems in measure theory, and counterexamples showing that their hypotheses are needed. $\endgroup$ – GEdgar Jan 1 '14 at 21:15
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    $\begingroup$ I edited the question so that it matches what you originally wrote, but I'm guessing you want $$\lim_{t\to t_1}E(X(t))$$ $\endgroup$ – Alex Becker Jan 1 '14 at 21:20
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In general, no.

Let $x(t,\omega) = \max(0, 2t(1-t\omega))$ for $\omega \in \Omega = (0,1]$. We have $\lim_{t \to \infty} x(t, \omega) = 0$ for all $\omega$, $x$ is continuous in $t$, $E [x(t,\cdot)] = 1$ for all $t$, but $E[\lim_{t \to \infty} x(t, \cdot)] = 0$.

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