Proofs of Theorems Concerning Convergence in Distribution

In this section we prove the equivalence of the statements in theorem 3A by showing that each implies its successor. For ease of writing, on occasion we write for for for , and for .

It is immediate that (i) implies (ii), since the function is a bounded continuous function of .

To prove that (ii) implies (iii), we make use of the basic formula (3.6) of Chapter 9 . For any define the function for any real number by

The function is continuous and integrable. Its Fourier transform is given for any by Therefore,

Thus we see that the Fourier transform is integrable. Consequently, from (3.6), of Chapter 9 we have.

By letting tend to in (5.3) and using the hypothesis of statement (ii), we obtain for any , as tends to , Next, define the function for any by

By the foregoing argument, one may prove that (5.4) holds for . Now, the expectations of the functions and clearly straddle the quantity : From (5.5), letting tend to , we obtain

Now, let tend to 0 in (5.6); since as tends to 0, it follows that (3.4) holds. Note that (5.7) would not hold if we did not require and to be points at which is continuous.

We next prove that (iii) implies (iv). Let be a positive number such that is continuous at and at . Then, for any real number Since statement (iii) holds, it follows that if is a continuity point of Now, also by (iii), since tends to , Consequently, which tends to , as one lets tend to . The proof that (iii) implies (iv) is complete.

We next prove that (iv) implies (i). We first note that a function , continuous on a closed interval, is uniformly continuous there; that is, for every positive number there is a positive number, derioted by , such that for any two points and in the interval satisfying. . Choose so that is continuous at and . On the closed interval is continuous. Fix , and let be defined as in the foregoing sentence. We may then choose real numbers having these properties: (i) , (ii) for , (iii) for is continuous at . Then define a function : It is clear that for Now where

Let be an upper bound for ; that is, for all . Then

Next, we may write as a sum of two integrals, one over the range and the other over the range . In view of (5.13), we then have Similarly In view of (5.14), (5.15), and the two preceding inequalities, it follows that

Letting first tend to 0 and then tend to , it follows that (5.13) will hold. The proof that (iv) implies (i) is complete.

The reader may easily verify that (v) is equivalent to the preceding statements.

Theoretical exercises