in terms of a definite integral as above. By the first part of fundamental theorem of calculus, Besides, Now, let x be any fixed real number, and let , which implies that , where is in the sense of definition 3. We have Here, the continuity of ln(y) is used, which follows from the continuity of 1/t: Here, the result lnan = nlna has been used. This result can be established for n a natural number by induction, or using integration by substitution. (The extension to real powers must wait until ln and exp have been established as inverses of each other, so that ab can be defined for real b as eb lna.) Characterization 1 ⇔ characterization 5 The following proof is a simplified version of the one in Hewitt and Stromberg, exercise 18.46. First, one proves that measurability (or here, Lebesgue-integrability) implies continuity for a non-zero function satisfying , and then one proves that continuity implies for some k, and finally implies . First, a few elementary properties from satisfying are proven, and the assumption that is not identically zero: If is nonzero anywhere (say at x=y), then it is non-zero everywhere. Proof: implies . . Proof: and is non-zero. . Proof: . If is continuous anywhere (say at x = y), then it is continuous everywhere. Proof: as by continuity at y. The second and third properties mean that it is sufficient to prove for positive x. If is a Lebesgue-integrable function, then It then follows that Since is nonzero, some can be chosen such that and solve for in the above expression. Therefore: The final expression must go to zero as since and is continuous. It follows that is continuous. Now, can be proven, for some k, for all positive rational numbers q. Let q=n/m for positive integers n and m. Then by elementary induction on n. Therefore, and thus for . If restricted to real-valued , then is everywhere positive and so k is real. Finally, by continuity, since for all rational x, it must be true for all real x since the closure of the rationals is the reals (that is, any real x can be written as the limit of a sequence of rationals). If then k = 1. This is equivalent to characterization 1 (or 2, or 3), depending on which equivalent definition of e one uses. Characterization 2 ⇔ characterization 4