Let, $T_n=\sum_{k=1}^{n}(X_k+Y_k)$. Then, $T_n/n\to EX_1+EY_1$ a.s.
By definition, $R_t=\sup\{n:T_{n-1}+X_n\leq t\}$
Then, $T(R_t-1)+X(R_t)\leq t< T(R_t)$ which implies, $\dfrac{T(R_t-1)+X(R_t)}{R_t}\leq \dfrac{t}{R_t}< \dfrac{T(R_t)}{R_t}$
After that, I don't understand how to get the answer. Thanks for any help.
I've been reading Rick Durret's book and i went through this problem.
Since no one have posted an answer i'll give it for completion in case if anyone is intrested in the future and search for this.
Let $Z_n=X_n+Y_n$. Then $Z_n$ corresponds to the total time that the system operated with $n$ light bulbs until the $n+1$ light bulb started working. Then $N_t=\sup\{n : Z_1+...+Z_n\leq t\}$ is the largest number of cycles that the system undergone from a working light bulb until the next light bulb started operating. Equivalently, $N_t$ also denote the number of light bulbs that we will have to change up to time $t$.
Now let $S_n=X_1+...+X_n$ denote the total time that $n$ light bulbs operate. Then, $R_t=S_{N_t}$ corresponds to the total amount of time that the light was working in the interval $[0,t]$. So, we need to find $\lim_{t\to \infty}R_t/t$.
Since, all $X_i$'s,$\ Y_i$'s are independent with each other and they have $F,G$ as distributions respectively, it follows that $Z_i$'s are independent and all have the same distribution (i.e. $F*G$ the convolution of $F,G$).
Now using the Renewal Theorem ( Example 2.4.1 in Durret ) we deduce that $$\tag{1}\lim_{t\to \infty}\frac{N_t}{t}=\frac{1}{EZ_1}=\frac{1}{EX_1+EY_1}$$ with probability 1, and $N_t\nearrow \infty$ as $t\to \infty$. Now, by the strong law of large numbers we also have $$\tag{2}\lim_{n\to \infty}\frac{S_n}{n}\to EX_1$$ again with probability 1. Now, using $(1),(2)$ and $N_t\to \infty$ as $t\to \infty$ with probability 1 we have $$\lim_{t\to \infty}\frac{R_t}{t}=\lim_{t\to \infty}\frac{S_{N_t}}{N_t}\cdot \frac{N_t}{t}=\frac{EX_1}{EX_1+EY_1}$$
Meaning that if we have $EY_1\ggΕΧ_1$ we will probably have to walk in the dark!