Let $f:[0,\infty) \times\mathbb{R}$ be bounded and continuous. If we let $(X_t)$ be an adapted process such that $X_0=0$ and
$X_t = \int_0^tf(s,X_s)ds \forall t \geq 0$
Show that $X_t$ is almost surely continuous.
We know that almost surely continuous means $ P(X_n \to X)=1 $ iff $$ \lim_{n \to \infty}P(\sup_{m \ge n} |X_m -X|>\epsilon) \to 0 $$
I feel like this would not be true.
Counter example:
Take f to be the constant function f = 1.
Then $X_t = \int_0^t1ds = t$ and the limit X=$\infty$
So $$ \lim_{n \to \infty}P(\sup_{t \ge n} |X_t -X|>\epsilon) $$
$$ =\lim_{n \to \infty}P(\sup_{t \ge n} |t -\infty|>\epsilon) $$
$$ =\lim_{n \to \infty}P(\infty>\epsilon) =1$$
So why is $X_t$ almost surely continuous and what is wrong with my counter example?