Over a month ago, I wrote about Steinhaus’ theorem, which states that if are sets of positive Lebesgue measure (we will use the notation to denote the Lebesgue measure of a set), then the set
contains an open interval. I want now to use the Steinhaus theorem to prove the uniqueness of a measurable solutions to the functional equation , where . Later, we will use this result to show that the only continuous probability distribution with the so-called “memorylessness” property is the exponential distribution.
We say that a function satisfies the Cauchy (functional) equation if
Note that the extension of to the domain is uniquely defined. Since is additive, we see that and therefore . We now prove the reverse direction. Clearly, , for . If and , then
The argument for the case , , and is completely analogous. Lastly, for the case ,
Lemma 1. If satisfies the Cauchy equation, then
Proof. Let and . Then
so dividing both sides by yields the case . Replacing by and applying additivity again yields the formula.
Observe that if is continuous, then by the density of in , we see that for all .
Lemma 2. If is additive, then in an interval , where , is bounded from above if and only if is bounded from below.
Proof. Suppose is bounded from above by . For , set . I claim that . Indeed,
Similarly, if is bounded from below by , then
Lemma 3. Suppose is additive and is bounded by in an interval . Then
In particular, is locally Lipschitz at and therefore continuous at .
Proof. For , let be as above. Since and , we see that and . Hence, it suffices to consider . I claim that we may choose a sequence such that , , and
Indeed, the function is continuous, so the claim follows from by the intermediate value theorem and the density of the rationals. Rewriting the above expression for , we see that
Analogously, we can choose a sequence such that , , and . Then
Lemma 4. If is additive and is bounded from above in an interval , where , then it is also bounded from above in the interval , where and .
Proof. Fix , and without loss of generality, we may assume that . For ,
where we know that the supremum is finite by hypothesis that is bounded on the interval .
Theorem 5. If satisfies the Cauchy functional equation and is measurable, then it is of the form
Proof. By a previous observation, it suffices to show that is continuous. Since , we see that for some . Set
so that for all , . By Steinhaus’s theorem, contains an open interval , where . In particular, is bounded on the interval , so by the preceding lemma, is bounded from above on the interval , where and . By Lemma 3, is continuous at .