Consider space $L^0$ of equivalent classes of measurable real random variables on a probability space $(\Omega,\mathcal F,P)$. Let $L_{+}^0=\{X\in L^0: X\geq 0\}$ and $L_{++}^0=\{X\in L^0: X>0\}$.
Consider the collection of balls $B(X,\epsilon):=\{Y\in L^0 : |X-Y|<\varepsilon\}$ with $X\in L^0$ and $\epsilon \in L_{++}^0$. This collection is a basis generating a topology on $L^0$.
I am reading a paper with this set-up, and at some point the author defines the following:
$$Y:=\lim_{n\to \infty} 1_{A_n}X$$
for some $X\in L^0$ and some increasing sequence $(A_n)\subset \mathcal F$.
Question: Why does this limit exists?
The limit exists a.s. by monotonicity, but it seems to me that the topology they have defined is not equivalent to a.s. convergence.
Am I missing something?
Your topology makes no mention of the measure $P$, so it certainly is not a.s. convergence, nor is it convergence in measure.
Among the positive functions $\varepsilon$ are the positive constants, so convergence in this topology implies uniform convergence. Taking $(0,1)$ with Borel sets (and Lebesgue measure), consider $X = 1$ and $A_n=(\frac{1}{n},1)$. Then $X_n := \mathbf1_{A_n}X$ does not converge uniformly, and thus does not converge in this topology.
What about "equivalence classes"? In this example, we have $P\big[|\mathbf1_{A_n}X - X| = 1 \big] > 0$ for all $n \in \mathbb N$, so taking $\varepsilon$ to be the constant $1/2$ we see that $$ \mathbf1_{A_n}X \not\in B\big(X,\varepsilon\big) $$ and $\mathbf1_{A_n}X$ does not converge to $X$.
We also claim that $\mathbf1_{A_n}X$ does not converge to any element $Y \in L^0$ not equal a.s. to $X$. Indeed: If $P[X \ne Y] > 0$ there is (constant) $\varepsilon \in L^0_{++}$ with $P\big[|X-Y|>\varepsilon\big] > 0$ and thus $P\big[|X_n-Y|>\varepsilon\big] > 0$ for large enough $n$.