Time average of a sample function is defined as:
$$\bar{x} = \langle~x(t)~\rangle = \lim \limits_{T \to \infty}\frac{1}{2T} \int \limits_{-T}^{T} x(t) ~ dt$$
This is how I see it: A few sample functions of X(t)=A, would be:
$$x_1(t) =0.2$$
$$x_2(t) = 0.7$$
$$\cdots$$
$$\text{etc}$$
How do they come up with 'A' as the time average??? $\langle x(t) \rangle=A$
I guess when you calculate the time average integral, you are suppose to treat r.v.'s as if they are constants:
$$\bar{x} = \lim \limits_{T \to \infty} \frac{1}{2T} \int \limits_{-T}^{T} x(t)~dt$$
$$\bar{x} = \lim \limits_{T \to \infty} \frac{1}{2T} \int \limits_{-T}^{T} A~dt$$
$$\bar{x} = A ~\lim \limits_{T \to \infty} \frac{1}{2T} \int \limits_{-T}^{T} 1~dt$$
$$\bar{x} = A ~\lim \limits_{T \to \infty} \frac{1}{2T} \bigg[t\bigg]_{-T}^{T} $$
$$\bar{x} = A ~\lim \limits_{T \to \infty} \frac{1}{2T} \bigg[T--T\bigg]$$
$$\bar{x} = A ~\lim \limits_{T \to \infty} \frac{2T}{2T}$$
$$\bar{x} = A ~\lim \limits_{T \to \infty} 1$$
$$\bar{x} = A$$