# Almost Surely discontinuous path

For $$t\in\mathbb{R}$$ let $$\{X_t\}$$ be a family of random variables independent and with the same distribution. Assume the distribution is absolutely continuous with respect to Lebesgue measure. I want to show that the paths of the process $$X_t$$ are discontinuous almost surely. Does anyone have a hint for this problem?

## 1 Answer

Lemma

Let $$(Y_n)$$ be i.i.d. If $$P(Y_n \to Y)>0$$ for some random variable $$Y$$ then $$Y$$ is degenerate.

Proof of the lemma: $$P(Y_{2n+1}-Y_{2n} \to 0)$$ is $$0$$ or $$1$$ by Kolmogorov's $$0-1$$ law. If it is $$1$$ then $$Y_{2n+1}-Y_{2n} \to 0$$ in distribution. Since this sequence is i.i.d. it follows that $$|\phi (t)|=1$$ for all $$t$$ where $$\phi$$ is the common characteristic function. This implies that $$Y_{2n+1}-Y_{2n}$$ is degenerate and hence $$Y_n$$ is degenerate for each $$n$$. The lemma follows.

Now consider $$P(X_{1/n} \to X_0)$$. If this probability is $$>0$$ then the lemma shows that $$X_0$$ has a degenerate distribution contradicting absolute continuity.