Convergence of sums of indicator random variables

I'm working through some practice problems for my final exam and I would like to get some ideas on tackling this problem:

Let $(\Omega,\mathcal{F})=(\mathbb{R}_+,\mathcal{B}(\mathbb{R}_+))$ and $\text{P}(d\omega)=\exp(-\omega)d\omega$ or equivalently,

$P[(a,b]]=\exp(-a)-\exp(-b)$ for $0\leq a\leq b < \infty$.

Let $X_n(\omega)=1_{[n,\infty)}(\omega)$ for $n\in\mathbb{N}$.

Does $S_n=\sum_{j=1}^n X_j$ converge almost surely to some limit S and does the partial sum $S_n$ converge to S in $L_1$?

My idea for the first part is to use Markov's inequality and Borel-Cantelli and show that $\sum P(|X_n|>\epsilon)<\infty$. I was able to find the probability distribution of $\mu_n(B)$ of $X_n$, for $\forall B\in\mathcal{B}$ to be:

P(X\in B)=\mu_n(B)=\left\{ \begin{aligned} P(\emptyset) = 0 && 0,1\not\in B\\ P\circ X^{-1}[0,n)=1-\exp(-n) && 0\in B, 1\not\in B\\ P\circ X^{-1}[n,\infty)=\exp(-n) && 0\not\in B, 1\in B\\ P\circ X^{-1}[0,\infty)=1 && 0,1\in B \end{aligned} \right.

For the second, I don't really have a clue except that i know that $\{X_n\}$ is uniformly integrable since it's uniformly bounded by by an $L_1$ random variable.

Edit: for later viewing

• By $P(\mathrm d\omega)=\exp(-\omega)\mathrm d\omega$ do you mean $P(B) = \int_B \exp(-\omega)\mathrm d\omega$ for $B\in\mathcal B(\mathbb R_+)$? – Math1000 Dec 14 '14 at 15:53
• Yes, it can be written equivalently as $P[(a,b]]=\exp(-a)-\exp(-b)$ for $0\leq a\leq b < \infty$. – stats134711 Dec 14 '14 at 15:58

Hint for the convergence in $L^1$ (since the almost sure convergence seems dealt with in the question): For every $n$, $$E\left(\sum_{k=n+1}^\infty X_k\right)=\frac{\mathrm e^{-n}}{\mathrm e-1},$$ and the RHS converges to zero when $n\to\infty$.
• Could you please tell me how you arrived at that quantity? I'm getting: $\sum E(S)$=$E(\sum X_k)$=$\sum E(X_k)$ = $\sum(0\cdot(1-e^{-k}) + 1\cdot e^{-k})=\sum_{k=n+1}^\infty (e^{-k})$=$\frac{e^{-(n+1)}}{1-e}$ – stats134711 Dec 14 '14 at 16:19