a question on 0-1 valued stochastic process Here's a question on probability theory from a layman (I'm a game theorist). It is very likely that the question will be a straightforward matter for someone who is a probability theorist. I guess I'm missing a very standard technique that could be used to address the problem. Any ideas would be very helpful! Thank you all very much in advance.
Consider a stochastic process $X_{t}$ taking values in the set $\{0,1\}$ according to the probability measure $\mu$.
Let 
$$Y_{t} = \mu\left(\limsup_{T \rightarrow \infty}\frac{1}{T}\sum_{t = 0}^{T - 1}X_{t} \geq \frac{1}{2}\mid X_{0},\dots,X_{t - 1}\right).$$
We know that
$$\mu(X_{t} = 1 \mid Y_{t} < \epsilon) > 1 - \epsilon.$$
We believe (but unable to prove) that 
$$Y_{t} = 1\text{ for every }t,\,\mu\text{-almost surely.}$$
This might have to do with Levy's zero-one law... 
Thanks again for your help!
P.S. My apologies for an earlier post that did not meet the standards of the forum.
 A: Let $A$ be the event $A = \left\{ \limsup_{T \to \infty}\frac{1}{T}\sum_{t = 0}^{T - 1}X_{t} \ge \frac{1}{2}\right\}$, so that $Y_t = P(A \mid X_1, \dots, X_t)$.  We will show that $P(A) = 1$ and hence $Y_t = 1$ almost surely for all $t$.
First, Lévy's zero-one law asserts that $Y_t \to 1_A$ almost surely.  Now assume there exists $\epsilon < 1/2$ such that for all $t$, we have $P(X_t = 1 \mid Y_t < \epsilon) > 1-\epsilon$ for all $t$ (I'm guessing that's what you want the quantifiers to be).  We can rewrite this as
$$P(X_t = 1, Y_t < \epsilon) > (1-\epsilon) P(Y_t < \epsilon) $$
or equivalently, letting $Z_t = 1_{\{Y_t < \epsilon\}}$,
$$E[X_t Z_t] > (1-\epsilon) E[Z_t].$$
Since $Y_t \to 1_A$ we have $Z_t \to 1_{A^c}$ almost surely, so by dominated convergence the right side converges to $(1-\epsilon)P(A^c)$.  On the left side, let us add and subtract $E[X_t 1_{A^c}]$:
$$E[X_t(Z_t - 1_{A^c})] + E[X_t 1_{A^c}] > (1-\epsilon)E[Z_t]$$
Since $Z_t \to 1_{A^c}$ almost surely, and $|X_t| \le 1$, the first term on the left goes to 0.  Now let us average over $T$:
$$\frac{1}{T} \sum_{t=1}^T E[X_t(Z_t - 1_{A^c})] + E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] > (1-\epsilon) \frac{1}{T} \sum_{t=1}^T  E[Z_t].$$
The first term on the left side goes to 0 and the right side goes to $(1-\epsilon) P(A^c)$, so we have
$$(1-\epsilon) P(A^c) \le \limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$
The quantity inside the expectation is bounded above by 1, so Fatou's lemma gives
$$\limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] \le E\left[\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$
But by definition of $A$, on the event $A^c$ we have $\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t \le \frac{1}{2}$.  Therefore
$$(1-\epsilon) P(A^c) \le \frac{1}{2} P(A^c).$$
Since $\epsilon < 1/2$ this implies $P(A^c) = 0$.
