Continuity at a point and empty interior

I am trying to solve the following problem:

Let $$f:[a, b] \rightarrow \mathbb{R}$$ be Riemann integrable. Suppose that $$f$$ is such that for any $$c \in [a,b]$$, $$f$$ continuous at $$c$$ implies $$f(c)=0$$. I need to prove that then the set $$X=\{x \in[a, b] ; f(x) \neq 0\}$$ has empty interior (notation: $$\operatorname{int}(x)=\varnothing$$).

What have i tried so far?

Since f is continuos at $$c \in [a, b]$$ for every $$\epsilon > 0$$ there exists $$\delta > 0$$ such that $$\forall$$ $$\bar{x} \in [a,b]$$ and $$|\bar{x} - c|<\delta$$ then $$|f(\bar{x})-f(c)|<\varepsilon$$

Since $$f(c) = 0$$ we have that $$|f(\bar{x})|<\varepsilon$$

Now i am stuck. Since i am trying to prove that $$X$$ has empty interior, i should come to the conclusion that there does not exist $$\bar{\epsilon} > 0$$ such that $$(x-\bar{\epsilon}, x+\bar{\epsilon}) \subset X$$ but i just cannot see the connection between what i have already done and the conclusion i should reach.

Can someone help?

• If it didn't have empty interior, then it would have positive measure, which would be a contradiction. Search "riemann integrable set of discontinuities" at MSE for related questions Dec 27, 2020 at 21:23
• I dont understand where the contradiction comes from. Can you explain it a little bit more? Also, do you think it is possible to provide a direct proof? Dec 27, 2020 at 21:57
• The set of poinst of discontinuity of a Riemann integrable function has measure zero. Dec 27, 2020 at 22:05
• But how is the set $X$ related to the points of discontinuity? Would you mind explaining it a little further, please? Dec 27, 2020 at 22:09
• Oh, i was missing something very simple. Since $f(x) \neq 0$, then $f$ is not continuos at $x$ (counter positive of the hypothesis) and therefore the relationship between the set of discontinuities and measure follows. Is that correct? Dec 27, 2020 at 22:11

$$1).\$$let $$D_{n}=\{x:\omega(f,x)\ge 1/n\}$$ and note that the points of discontinuity of $$f$$ are given by $$\bigcup D_n.$$
$$2).\ f\$$ is integrable so we may take a partition $$I_j$$ such that $$\sum \omega(f,I_j)|I_j|<\epsilon/n$$ and let $$\Delta$$ be the set of indices $$j$$ such that $$D_{1/n}$$ has non-empty intersection with the interior of the $$I_j.$$
$$3).\$$ Conclude that $$D_{1/n}$$ is contained in $$\bigcup_{j\in \Delta} I^\circ_j$$ plus possibly the endpoints of these intervals.
$$4).\$$ Conclude that $$\sum_{j\in \Delta}|I_j|<\epsilon$$ and therefore that $$D_{1/n}$$ can be covered by finitely many $$\textit{open}\$$intervals the sum of whose lengths is less than $$2\epsilon$$ (the factor of $$2$$ enters because you have to consider the endpoints of the $$I_j$$, which is no problem since the number of these is finite).
$$5).\$$ Conclude now that as $$\epsilon$$ is arbitrary, $$D_{1/n}$$ and hence $$D$$ cannot contain an open interval-