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If $X_t, t \in T$ is separable random process (with values in metric space), an event that it's sample path be uniformly continuous on $T$ $$ \bigcap_{m \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \bigcap_{t_1 \in T} \bigcap_{t_2 \in K (t_1, 1 / n)} \left\{ d (X_{t_1}, X_{t_2}) \leqslant 1 / m \right\} $$ differs from obviously measurable set $$ \bigcap_{m \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \bigcap_{t_1 \in D} \bigcap_{t_2 \in K (t_1, 1 / n) \cap D} \left\{ d (X_{t_1}, X_{t_2}) \leqslant 1 / m \right\} $$ only in set of measure zero. Does this also holds for an event that it's sample path be merely continuous on $T$ $$ \bigcap_{t_1 \in T} \bigcap_{m \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \bigcap_{t_2 \in K (t_1, 1 / n)} \left\{ d (X_{t_1}, X_{t_2}) \leqslant 1 / m \right\} $$ I have been told that this is the case, but since I can no longer exchange intersections $\bigcap_{t_1 \in T}$and $\bigcap_{t_2 \in K (t_1, 1 / n)}$, I don't see how would one prove this.

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What is $T$? If it's a locally compact space, a function from $T$ to a metric space is continuous if and only if it is uniformly continuous on all compact subsets of $T$. So if $T$ is $\sigma$-compact, the event "$X$ is continuous" is the intersection of countably many events "$X$ is uniformly continuous on $T_n$".

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