Let's denote the measure space by $(X, \mathcal{B}, \mu)$ and the measure-preserving transformation by $T: X \to X$. Let $A \in \mathcal{B}$ be a measurable set with $0 < \mu(A) < \infty$. Let $0 < t < 1$ and $n \ge 1$. For any $s > 0$, we set $Y_s := \left\{x \in X: \sum_{i=1}^{n} 1_A \circ T^i(x) \ge s\right\}.$
We aim to show:
$$\frac{1}{n+1} \int_X \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{(1-t)\mu(A)}{1 + \sup\left\{s > 0: \mu\left(A \cap Y_s\right) \ge t \mu(A)\right\}}.$$
Approach:
Let us assume $s^* = \sup\left\{s > 0: \mu\left(A \cap Y_s\right) \ge t \mu(A)\right\}$. We aim to show that:
$$\frac{1}{n+1} \int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{(1-t)\mu(A)}{1 + s^*}.$$
Now, if we take $f = \sum_{i=1}^{n} 1_A \circ T^i$, then we have $\frac{f}{n+1} \le \max_{0 \le k \le n} 1_A \circ T^k$. It follows that:
$$\frac{1}{n+1} \int_A f \, d\mu \le \int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu.$$
By the definition of $Y_s$,
$$Y_s = \left\{x \in X: \sum_{i=1}^{n} 1_A \circ T^i(x) \ge s\right\}.$$and since $s^*$ is the supremum, we have,
$$\int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge \int_{A \cap Y_{s^*}} s^* \, d\mu \ge s^* \cdot \mu(A \cap Y_{s^*}).$$
By our assumption, $\mu(A \cap Y_{s^*}) \ge t \mu(A)$. Hence,
$$\int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge s^* \cdot t \cdot \mu(A).$$
Therefore,
$$\frac{1}{n+1} \int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge \frac{s^* \cdot t \cdot \mu(A)}{n+1}.$$
Combining the results, we obtain:
$$\int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{s^* \cdot t \cdot \mu(A)}{n+1}.$$At this point, I am unsure how to proceed further. Could you please assist me in solving the remaining steps? I appreciate your time and effort. Thank you.