From Example 4 of Section 2.3 of Apostol, we know that given a nonnegative, integrable function on an interval , and the area of the ordinate set of is , then if we define a function , we have,

Generalize this formula and prove the result.

To generalize this, we proceed as follows.

Let be a nonnegative integrable function on , and let be the ordinate set of . If we apply a transformation under which we multiply the -coordinate of each point on the graph of by a constant and each -coordinate by a constant , then we obtain a new function were a point is on if and only if is on . Then,

Letting denote the ordinate set of :

I think it would be better to prove that the ordinate set of the new graph is measurable first. To do this, the solution (the part from a(S) to jk . a(S)) only needed to be written in a reverse order

The integration interval Is [ka,jb].

No, [ka, kb] is correct. The horizontal scale factor is k. The interval [ka, kb] is the horizontal dimension (along the x-axis) of the ordinate set of g, after the ordinate set of f is scaled horizontally by k and vertically by j.

Richard, why do you assume that “the horizontal scale factor is k”? The problem statement says nothing about scaling [a, b] by the same factor in both ends, and you could perfectly scale the left-end by ka and the right end by jb, yielding the interval [ka, jb].

The integration interval for the function G (x) is wrong … it is not [ka, kb]