Froda's theorem

In mathematics, Darboux–Froda's theorem, named after Alexandru Froda, a Romanian mathematician, describes the set of discontinuities of a monotone real-valued function of a real variable. Usually, this theorem appears in literature without a name. It was written in Froda' thesis in 1929.. As it is acknowledged in the thesis, the theorem is in fact due to Jean Gaston Darboux.

Definitions

1. Consider a function f of real variable x with real values defined in a neighborhood of a point $x_0$ and the function f is discontinuous at the point on the real axis $x=x_0$. We will call a removable discontinuity or a jump discontinuity a discontinuity of the first kind.
2. Denote $f(x+0):=\underset{h\searrow 0}{\lim }f(x+h)$ and $f(x-0):=\underset{h\searrow 0}{\lim }f(x-h)$. Then if $f(x_0+0)$ and $f(x_0-0)$ are finite we will call the difference $f(x_0+0)-f(x_0-0)$ the jump of f at $x_0$.

If the function is continuous at $x_0$ then the jump at $x_0$ is zero. Moreover, if $f$ is not continuous at $x_0$, the jump can be zero at $x_0$ if $f(x_0+0)=f(x_0-0)\ne f(x_0)$.

Precise statement

Let f be a real-valued monotone function defined on an interval I. Then the set of discontinuities of the first kind is at most countable.

One can prove that all points of discontinuity of a monotone real-valued function defined on an interval are jump discontinuities and hence, by our definition, of the first kind. With this remark Froda's theorem takes the stronger form:

Let f be a monotone function defined on an interval $I$. Then the set of discontinuities is at most countable.

Proof

Let $I:=[a,b]$ be an interval and $f$, defined on $I$, an increasing function. We have

$f(a)\le f(a+0)\le f(x-0)\le f(x+0)\le f(b-0)\le f(b)$

for any $a<x<b$. Let $\alpha >0$ and let $x_1<x_2<\cdots <x_n$ be $n$ points inside $I$ at which the jump of $f$ is greater or equal to $\alpha$:

$f(x_i+0)-f(x_i-0)\ge \alpha ,i=1,2,\dots ,n$

We have $f(x_i+0)\le f(x_{i+1}-0)$ or $f(x_{i+1}-0)-f(x_i+0)\ge 0,i=1,2,\dots ,n$. Then

$f(b)-f(a)\ge f(x_n+0)-f(x_1-0)=\sum _{i=1}^n[f(x_i+0)-f(x_i-0)]+$

$+\sum _{i=1}^{n-1}[f(x_{i+1}-0)-f(x_i+0)]\ge \sum _{i=1}^n[f(x_i+0)-f(x_i-0)]\ge n\alpha$

and hence: $n\le \frac{f(b)-f(a)}{\alpha }$.

Since $f(b)-f(a)<\mathrm{\infty }$ we have that the number of points at which the jump is greater than $\alpha$ is finite or zero.

We define the following sets:

$S_1:=\{x:x\in I,f(x+0)-f(x-0)\ge 1\}$,

$S_n:=\{x:x\in I,\frac{1}{n}\le f(x+0)-f(x-0)<\frac{1}{n-1}\},n\ge 2.$

We have that each set $S_n$ is finite or the empty set. The union $S=\bigcup _{n=1}^{\mathrm{\infty }}{S}_n$ contains all points at which the jump is positive and hence contains all points of discontinuity. Since every $S_i,i=1,2,\dots$ is at most countable, we have that $S$ is at most countable.

If $f$ is decreasing the proof is similar.

If the interval $I$ is not closed and bounded (and hence by Heine–Borel theorem not compact) then the interval can be written as a countable union of closed and bounded intervals $I_n$ with the property that any two consecutive intervals have an endpoint in common: $I=\underset{n=1}{\overset{\mathrm{\infty }}{\cup }}{I}_n.$

If $I=(a,b],a\ge -\mathrm{\infty }$ then $I_1=[{\alpha }_1,b],I_2=[{\alpha }_2,{\alpha }_1],\dots ,I_n=[{\alpha }_n,{\alpha }_{n-1}],\dots$ where $\{{\alpha }_n{\}}_n$ is a strictly decreasing sequence such that ${\alpha }_n\to a.$ In a similar way if $I=[a,b),b\le +\mathrm{\infty }$ or if $I=(a,b)-\mathrm{\infty }\le a<b\le \mathrm{\infty }$.

In any interval $I_n$ we have at most countable many points of discontinuity, and since a countable union of at most countable sets is at most countable, it follows that the set of all discontinuities is at most countable.

See also

• Continuous function
• Classification of discontinuities

References

• Bernard R. Gelbaum, John M. H. Olmsted, Counterexamples in Analysis, Holden–Day, Inc., 1964. (18. Page 28)
• John M. H. Olmsted, Real Variables, Appleton–Century–Crofts, Inc., New York (1956), (Page 59, Ex. 29).