Exchanging expectation and limits Exchanging expectation and limits
I have a stochastic process, 
${b_t}
\, (t=0, 1, 2, \ldots)$, which follows a random walk. Specifically, 
${b_0} = 0$
 and for $t$ greater than zero,
$\displaystyle {b_t} = \sum\limits_{i = 1}^t {{\varepsilon _i}} $.
The ${\varepsilon _i}$
 are i.i.d. random variables with zero expected value. Other than the existence of the expectation, I would rather not have to assume any stronger regularity on the
 ${\varepsilon _i}$.
Let 
$0 < \beta  < 1$
 (beta is a discount factor).
My question is: 
 Is it the following true?
$${E_{t = 0}}\left( {\mathop {\lim }\limits_{t \to \infty } {\beta ^t}{b_t}} \right) = 0$$ 
By ${E_{t = 0}}$ I mean the expected value at time zero. It seems to me that the proof depends on whether I can exchange the limit and the expectation, since 
$\forall t, {E_{t = 0}}\left( {{b_t}} \right) = 0.$
 Please, provide the details of the proposed proof, since I am not a mathematician (just an economist). Thank you.
Thank you!
By working in the first two answers received I realized that the nature of my problem (stating a meaningful transversality condition for an optimization problem, and proving that the proposed solution meets it) required me to prove that 
${\beta ^t}{b_t}\mathop  \to \limits^{a.s.} 0$
 and not only that ${E_{t = 0}}\left( {\mathop {\lim }\limits_{t \to \infty } {\beta ^t}{b_t}} \right) = 0$.
I think that the third answer, based on the Strong Law of Large Numbers, kills both birds with the same stone, and a clear and elegant stone to boot. Thanks to all.
 A: Assuming that $Var(\epsilon_1)=\sigma_{\epsilon}^2$ we have
$$\mathbb{E}\beta^{2t}b_t^2= \sigma_{\epsilon}^2\frac{\beta^2}{(\beta^2-1)^2}+o(1)$$
Hence, $\beta^tb_t$ is $L^2$-bounded and uniformly integrable so that you can pass the limit inside the expectation... 
Edit:
In the first step, you need to show that $\beta^tb_t$ converges a.s. (to $0$). This can be done in various ways, e.g. using Kolmogorov's two-series theorem or the SLLN.
A: By Strong Law of Large Numbers, without any assumptions on the variance of $\epsilon_i$,
$$
\frac{1}{n}\sum^n_{i=1}\epsilon_i \rightarrow \mathbb E[\epsilon_i]=0 \quad \text{almost surely}
$$
So since $\lim_{n \rightarrow \infty} n\beta^n = 0$ and $n^{-1}b_n \rightarrow 0$ almost surely, we have that
$$
 \beta^n b_n\rightarrow 0 \quad \text{almost surely}
$$
Now
$$
\mathbb E \left[\lim_{n\rightarrow \infty} \beta^n b_n\right] = \mathbb E [0]=0
$$
since $\lim_{n\rightarrow \infty} \beta^n b_n=0$ on the whole probability space except on a set of measure $0$. 
So this is done without any interchanging of limits and expectations.  
