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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.

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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

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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