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?

Thanks in advance, Lucas

  • 1
    $\begingroup$ 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 $\endgroup$
    – Mirko
    Dec 27, 2020 at 21:23
  • $\begingroup$ 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? $\endgroup$
    – Lucas
    Dec 27, 2020 at 21:57
  • 1
    $\begingroup$ The set of poinst of discontinuity of a Riemann integrable function has measure zero. $\endgroup$
    – copper.hat
    Dec 27, 2020 at 22:05
  • $\begingroup$ But how is the set $X$ related to the points of discontinuity? Would you mind explaining it a little further, please? $\endgroup$
    – Lucas
    Dec 27, 2020 at 22:09
  • $\begingroup$ 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? $\endgroup$
    – Lucas
    Dec 27, 2020 at 22:11

1 Answer 1


Hints for a proof from scratch, using your idea:

$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-


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