# Why are the two processes in this example not indistinguishable?

This is taken from example 1.4 in Brownian Motion and Stochastic calculus, by Karatzas and Shreve. There's obviously something I've misunderstood, but here goes:

Assume that $$T$$ is a positive random variable with a continuous distribution. Let $$(X_t)_{t\ge 0},(Y_t)_{t\ge 0}$$ be two stochastic processes where $$X_t = 0\,\, \forall t\ge 0$$ and

$$Y_t = \begin{cases} 0, & t \neq T, \\ 1, & t = T. \end{cases}$$

Then $$X$$ and $$Y$$ are modifications of each other since

$$\forall t\ge 0\colon \, \mathbb{P}(X_t = Y_t) = \mathbb{P}(t \neq T) = 1.$$

However, $$\mathbb{P}(\forall t\ge 0\colon X_t = Y_t) = 0$$, so the processes aren't indistinguishable.

My problem: There is a result that says if $$X$$ and $$Y$$ are modifications and have $$\mathbb{P}$$-a.s. right-continuous sample paths, then $$X$$ and $$Y$$ are indistinguishable.

Here, $$t \mapsto X_t$$ is constant, so it must be continuous, and $$t \mapsto Y_t$$ is discontinuous only on a null set, so it must be $$\mathbb{P}$$-a.s. continuous.

Hence $$X,Y$$ should be indistinguishable, no?

It is true that the paths of $$Y$$ all have only one point of discontinuity (namely the value of $$T$$), but that still means that all paths of $$Y$$ are not (right-)continuous, so in fact $$Y$$ is right-continuous with probability $$0$$, which is why the result you cited does not apply here.
You were confusing null sets $$B\subset \mathbb{R}$$ with respect to Lebesgue measure and null sets $$N\subset \Omega$$ with respect to the probability measure $$\mathbb{P}$$ (where $$\mathbb{P}$$ is defined on some measurable space $$(\Omega,\mathcal{A})$$).