Let P, Q and f, g be as in Exercise 3 and, without loss of generality, we suppose that the dominating measure is a

Let P, Q and f, g be as in Exercise 3 and, without loss of generality, we suppose that the dominating measure µ is a probability measure (e.g,. µ = (P + Q)/2). Define ( by ( = ( ( (fg)1/2 d µ, and show that:
(i) ( ( 1.
(ii) 2(1 - () ( d(P, Q) ( 2(1 - (2)1/2, where (by Exercise 3), d(P, Q) = ((| ( - g|dµ = || P - Q}||.
Replacing P, Q and f, g, ( by Pn, Qn and (n, gn, (n, n ( 1, part (ii) becomes:
2(1 - pn) d (Pn, Qn) ( 2(1 - (2n)1/2.
(iii) From this last relation, conclude that, as n → (, d(Pn, Qn) → 0 if and only if (n → 0.