Prove a.s. convergence of $(X_n)_n$ satisfying $E(X_{n+1} \mid F_n) \leq X_n+Y_n$ for $\sum_n Y_n<\infty$ I am have a problem with proving a convergence of a sequence of random variables in the given context:

Let $F_n$ be a filtration. Assume that $X_n$ and $Y_n$ are non-negative and integrable $F_n$-adapted random variables for $n\geq 0$.
Furthermore, I assume that:
$\mathbb{E}[X_{n+1} \mid F_n]\leq X_n+Y_n$ and that  $\sum_{n}Y_n < \infty$.
Under the above conditions, I should show that $X_n$ converges almost surely for $n\rightarrow \infty$.

My approach has been to define a variable $Z_n=X_n-\sum_{i=0}^{n-1} Y_i$, and to set this into the expression for the conditional mean of $X_{n+1}$, but this did not gave any result.
Anybody has an idea to this?
 A: It follows from your assumption on the conditial expectation that
$$Z_n = X_n - \sum_{i=0}^{n-1} Y_i$$
is a supermartingale. If we define
$$T_k := \inf\left\{n \in \mathbb{N}; \sum_{i=0}^n Y_i \geq k\right\}$$
for fixed $k \in \mathbb{N}$, then $T_k$ is an $F_n$-stopping time. By the optional stopping theorem, the stopped process $(Z_{n \wedge T_k})_{n \in \mathbb{N}}$ is also a supermartingale. Moreover, the definition of $T_k$ and the non-negativity of $X_n$ and $Y_n$ entail that $Z_{n \wedge T_k} \geq -k$ for all $n \in \mathbb{N}$, and therefore
$$\sup_{n \in \mathbb{N}} \mathbb{E}(Z_{n \wedge T_k}^-) < \infty.$$
Applying the standard convergence theorem for supermartingales, we conclude that the limit $\lim_{n \to \infty} Z_{n \wedge T_k}$ exists almost surely, and so
$$\Omega_0 := \bigcap_{k \geq 1} \{\omega \in \Omega; \lim_{n \to \infty} Z_{n \wedge T_k}(\omega) \, \, \text{exists}\}$$
has probability $1$. Now if $\omega \in \Omega_0$, then $\sum_{n} Y_n(\omega)<\infty$ implies that we can choose $k \in \mathbb{N}$ large enough such that $T_k(\omega)=\infty$. As $\omega \in \Omega_0$ we thus know that
$$\lim_{n \to \infty} Z_{n \wedge T_k}(\omega) = \lim_{n \to \infty} Z_n(\omega) = \lim_{n \to \infty} \left( X_n(\omega)- \sum_{i=0}^{n-1} Y_i(\omega) \right)$$
exists. Using once more that $\sum_{i} Y_i(\omega)<\infty$, this shows that $\lim_n X_n(\omega)$ exists.
