I think the answer is yes. First replace the open intervals with closed intervals, contained in the open ones. This makes it harder. Then let $ A_N = \alpha \in (0,1] \; \text{, such that } f(n \alpha) = 0 \forall n > N $ where f is the indicator function of the set. $A_N$ is closed since if $\alpha_i \rightarrow \alpha, m\alpha_i \rightarrow m \alpha$ and since all of the $m\alpha_i$ are in the union of the (closed) intervals, $m\alpha$ is also. Also, $[0,1] = \cup 0, A_i$, if the conjecture is false, so under those circs, by the Baire category thm, one of the $A_i$ has non-empty interior. Once that has been achieved it is straightforward to show $f(x) \rightarrow 0$ . That is a contradiction. This is based on a known thm about cont functions, that if they converge along every sequence, they converge. I saw it in the math monthly roughly 1984 , where it was called a 'folk theorem'.