# an integrable function can be discontinuous in all of the points of an interval?

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

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

• 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 $D_m$? – Majid May 16 at 1:17
• 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? – RRL May 16 at 2:20
• @majid: Your second doubt is valid. I think I can fix the argument when I get a chance. – RRL May 16 at 7:39
• Thanks for the explanation – Majid May 16 at 14:56
• @Majid: I fixed the proof. This makes no mention of Lebesgue measure. – 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.

• 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? – Majid May 14 at 21:52
• Right now, I don't see any other way of doing it. – José Carlos Santos May 14 at 21:55
• 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. – DanielWainfleet May 15 at 19:05