Let $\{X_i\}$ be a sequence of i.i.d. random variables, $\overline X_n = \displaystyle\frac{1}{n}\sum_{i=1}^n X_i$ be their $n$th partial mean, and $\mu = \mathbb E[X_1] < \infty$.
Let $x \in \mathrm{supp}(X_1)$. Under which conditions does $$ \overline X_n \mid \left\{\overline X_n \ge x \right\} \to \begin{cases} \mu, &\text{if } x \le \mu\\ x, &\text{if } x > \mu \end{cases} $$ converge almost surely or in probability?
Wikipedia page says that the LLN holds if the random variables are exchangeable instead of mutually independent. Thus, I imagine that the above is true almost surely, but something seems off to me.
The case $x \le \mu$ is okay because $\overline X_n \to \mu \ge x$ almost surely, thus the condition will not affect the convergence.
The case $x > \mu$ is the interesting one. Assuming finite variance $\sigma^2$, then $Z_n = \sqrt{n}\frac{\overline X_n - \mu}{\sigma}$ converges in distribution to $\mathcal N(0,1)$ and \begin{align*} \mathbb P\left(\overline X_n - x > \varepsilon \mid \overline X_n \ge x\right) &= \mathbb P\left(Z_n > \sqrt{n}\,\frac{x - \mu + \varepsilon}{\sigma} \mid Z_n \ge \sqrt{n}\,\frac{x - \mu}{\sigma}\right) \\ &\to \lim_{n\to\infty}\frac{\text{erfc}\left(\frac{\sqrt{n} (x -\mu +\epsilon )}{\sqrt{2} \sigma }\right)}{\text{erfc}\left(\frac{\sqrt{n} (x-\mu )}{\sqrt{2} \sigma }\right)} = 0 \end{align*} for every $\varepsilon > 0$. Thus, for finite variance, $\overline X_n \mid \left\{\overline X_n \ge x \right\} \to x > \mu$ in probability.