# 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.[dubious ]. 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 $f(x+0):=\lim _{h\searrow 0}f(x+h)$  and $f(x-0):=\lim _{h\searrow 0}f(x-h)$ . 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 $f(x_{0}+0)=f(x_{0}-0)\neq f(x_{0})$ .

## 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

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

for any $a . Let $\alpha >0$  and let $x_{1}  be $n$  points inside $I$  at which the jump of $f$  is greater or equal to $\alpha$ :

$f(x_{i}+0)-f(x_{i}-0)\geq \alpha ,\ i=1,2,\ldots ,n$

We have $f(x_{i}+0)\leq f(x_{i+1}-0)$  or $f(x_{i+1}-0)-f(x_{i}+0)\geq 0,\ i=1,2,\ldots ,n$ . Then

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

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

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

We define the following sets:

$S_{1}:=\{x:x\in I,f(x+0)-f(x-0)\geq 1\}$ ,
$S_{n}:=\{x:x\in I,{\frac {1}{n}}\leq f(x+0)-f(x-0)<{\frac {1}{n-1}}\},\ n\geq 2.$

We have that each set $S_{n}$  is finite or the empty set. The union $S=\bigcup _{n=1}^{\infty }S_{n}$  contains all points at which the jump is positive and hence contains all points of discontinuity. Since every $S_{i},\ i=1,2,\ldots$  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: $I=\cup _{n=1}^{\infty }I_{n}.$

If $I=(a,b],\ a\geq -\infty$  then $I_{1}=[\alpha _{1},b],\ I_{2}=[\alpha _{2},\alpha _{1}],\ldots ,\ I_{n}=[\alpha _{n},\alpha _{n-1}],\ldots$  where $\{\alpha _{n}\}_{n}$  is a strictly decreasing sequence such that $\alpha _{n}\rightarrow a.$  In a similar way if $I=[a,b),\ b\leq +\infty$  or if $I=(a,b)\ -\infty \leq a .

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.