We have a random 0-1 sequence $(Y_k)_{k\in\mathbb{N}}$ (only takes values from $\{0,1\}$). We do not assume $Y_0, Y_1,\ldots$ are i.i.d. (even not independent). Then we are interested whether the following equality is true: $$ \limsup_{k\to\infty} \frac{1}{k} \sum_{i=0}^{k-1}Y_i = \limsup_{k\to\infty} \frac{1}{k}\sum_{i=0}^{k-1}\mathbb{E}\left(Y_i\mid \mathcal{F}_{i-1} \right), $$ where $\mathcal{F}_{k-1}$ is the $\sigma$-algebra generated by $Y_0,\ldots,Y_{k-1}$.
So far, we know that the equality is true when $Y_0, Y_1,\ldots$ are independent by Kolmogorov Convergence Criterion.
Any comment helps.
Let $d_i:=Y_i-\mathbb E\left(Y_i\mid\mathcal F_{i-1}\right)$. It suffices to prove that $$\lim_{k\to\infty}\frac 1k\sum_{i=1}^k d_i=0 \mbox{ a.s.}.$$ To do so, we can check that for each positive $\varepsilon$, the series $\sum_{k=1}^\infty \mathbb P\left(\left\lvert \sum_{i=1}^kd_i\right\rvert>k\varepsilon\right)$ converges, which is a consequence of Azuma-Hoeffding's inequality.
Note that we only used the fact that $Y_i$ is bounded by $1$. The wanted equality of limsups certainly holds under weaker assumptions.