Let $X\_t$ be a sequence of i.i.d. random variables with mean $\mu$. Then the law of large numbers states that
$$\lim\_{T \to \infty} \frac1T \sum\_{t=1}^T X\_t = \mu \quad a.s.$$ 


Now suppose that (in a game theoretic context) an agent can choose at every instant of time if she wants to observe $X\_t$ or not. I want to prove that the average over the observations still converges to $\mu$. 


In more details, the let $k\_t=1$ denote that $X\_t$ is observed, and $k\_t=0$ that $X\_t$ is not observed. To model that the agent's choice at time $t$ can depend only on past observations, I require $k\_t$ to be measurable with respect to the $\sigma$-algebra $$\mathcal F\_{t-1} := \sigma(k\_1 X\_1, \ldots, k\_{t-1} X\_{t-1}).$$ I define
$$N\_T:=\sum\_{t=1}^T k\_t, \quad Y\_T:=\sum\_{t=1}^T k\_t X\_t$$ and assume that $N\_T \to \infty$ as $T \to \infty$. Now the question is if 
$$\lim\_{T \to \infty} \frac{1}{N\_T}Y\_T = \mu.$$



Law of large numbers for stochastically chosen samples
Using John's notation, and assuming $\{X\_{\sigma(i)}\}$ are independent, then $X=(X\_1,X\_2,\ldots)$ has the same distribution as $X\_\sigma=(X\_{\sigma(1)},X\_{\sigma(2)},\ldots)$. Let $f(X)=\limsup\_{n\rightarrow\infty}(X\_1+\cdots+X\_n)/n$. Then $f(X\_\sigma)=\limsup\_{t\rightarrow\infty}Y\_t/N\_t$, and $\mathbb P[f(X\_\sigma)=\mu]=\mathbb P[f(X)=\mu]=1$, and similarly for the liminf.


