I am trying to solve the following question:

$f:[a,b]\to\mathbb{R}$ is integrable. Also if $f$ is continuous in a point $c\in [a,b]$ then $f(c)=0$.

Show that $X=\{x\in [a,b] : f(x)\neq 0\}$ has empty interior.

My attempt: Assuming that $X$ has nonempty interior, $\forall x\in X$, there exists some $\delta$ such that $(x-\delta,x+\delta)\subset X$. Now by hypothesis, $f$ is not continuous on $(x-\delta,x+\delta)$. One can get any contradiction according to the fact that $f$ is integrable?

If someone can offer any other solution will be thankful.


Since $f$ is integrable, for any $\epsilon > 0$ there exists a partition $P = (x_0,x_1, \ldots,x_m)$ of $[a,b]$ such that the difference of upper and lower Darboux sums satisfies

$$U(P,f) - L(P,f) = \sum_{j=1}^m \omega(f,[x_{j-1},x_j])(x_j - x_{j-1}) < \epsilon,$$

where $ \omega(f,[x_{j-1},x_j])= \sup_{x,y \in [x_{j-1},x_j]} |f(x) - f(y)|$ is the oscillation of $f$ on the interval $[x_{j-1},x_j]$.

Let $\omega_f(x) = \lim_{\delta \to 0}\omega(f,[x- \delta,x+\delta])$ denote the oscillation at a point $x$. Since $f$ is continuous at $x$ if and only if $\omega_f(x) = 0$, the set of discontinuity points is given by

$$D_f = \bigcup_{k=1}^\infty D_k =\bigcup_{k=1}^\infty \left\{x\in [a,b]:\omega_f(x) \geqslant \frac{1}{k} \right\}$$

Since $f(x)=0$ at any point in $[a,b]$ where $f$ is continuous, we have $X = D_f$.

Suppose $X$ has nonempty interior. By the Baire category theorem, one of the sets $D_m$ must have nonempty interior and there exists an interval $[\alpha, \beta]\subset D_m$. We have $\omega_f(x) \geqslant 1/m$ for all $x \in [\alpha, \beta]$ and $\omega(f,I) \geqslant 1/m$ for any interval $I$ that intersects $[\alpha,\beta]$.

Choosing $\epsilon < (\beta - \alpha)/m$, we get a contradiction

$$\epsilon > \sum_{j=1}^m \omega(f,[x_{j-1},x_j])(x_j - x_{j-1}) \geqslant \sum_{[x_{j-1},x_j]\cap [\alpha,\beta]\neq \emptyset} \omega(f,[x_{j-1},x_j])(x_j - x_{j-1})\geqslant \frac{\beta-\alpha}{m}$$

  • $\begingroup$ it sounds a nice answer. .But I have some doubts. First why there exists a compact interval contained in X and second why $[\alpha,\beta]$ is contained in some $D_m$? $\endgroup$ – Majid May 16 at 1:17
  • 1
    $\begingroup$ If $X$ has non-empty interior it must contain an open interval $(\alpha',\beta')$ correct? And if that is true what can we find inside of that open interval? $\endgroup$ – RRL May 16 at 2:20
  • $\begingroup$ @majid: Your second doubt is valid. I think I can fix the argument when I get a chance. $\endgroup$ – RRL May 16 at 7:39
  • $\begingroup$ Thanks for the explanation $\endgroup$ – Majid May 16 at 14:56
  • $\begingroup$ @Majid: I fixed the proof. This makes no mention of Lebesgue measure. $\endgroup$ – RRL May 18 at 4:35

The set of discontinuity points of a Riemann-integrable function $f$ from an interval $[a,b]$ into $\mathbb R$ has Lebesgue measure $0$. And a set with Lebesgue measure $0$ has an empty interior.

  • $\begingroup$ thanks @Joce Carlos Santos for quick answer. There in any other solution that does not use measure theory as it is an exercise related to Real Analysis? $\endgroup$ – Majid May 14 at 21:52
  • 1
    $\begingroup$ Right now, I don't see any other way of doing it. $\endgroup$ – José Carlos Santos May 14 at 21:55
  • $\begingroup$ Nothing wrong with this A, but having empty interior is a weaker condition than being Lebesgue-null and there is now another A that addresses the weaker condition directly without measure theory. $\endgroup$ – DanielWainfleet May 15 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.